Permutations

i. Given

= 30 × 29

= 870

ii. Given

We know that:

11! = 11 × 10 × 9 × … × 1

10! = 10 × 9 × 8 × … × 1

9! = 9 × 8 × 7 × … × 1

Putting these values, we get,

= 110 – 10

Hence,

iii. We have to find LCM of 6!, 7!, and 8!

We can write as,

= 8! = 8 × 7 × 6!

= 7! = 7 × 6!

= 6! = 6!

Therefore,

L.C.M (6!, 7!, 8!) = LCM [8 × 7 × 6!, 7 × 6!, 6! ]

= 8 × 7 × 6!

Hence, LCM is 8!

Given:

L.H.S

Hence, L.H.S = R.H.S

We have

We know that

5! = 5 × 4 × 3 × 2 × 1

6! = 6 × 5 × 4 × 3 × 2 × 1

Hence, x = 36

We have

We can write as

10! = 10 × 9!

9! = 9 × 8!

By putting these values

Hence, x = 100

We have

We can write as.

8! = 8 × 7 × 6!

7! = 7 × 6!

Putting the value, we get

Hence, x = 64

Given 5 . 6 . 7 . 8 . 9 . 10

Find: Convert into Factorial

Now, We can write as

=

Hence,

Given 3 . 6 . 9 . 12 . 15 . 18

Find: Convert into Factorial

= (3 × 1) × (3 × 2) × (3 × 3) × (3 × 4) × (3 × 5) × (3 × 6)

= 3⁶ (1 × 2 × 3 × 4 × 5 × 6)

= 3⁶ (6!)

Given (n + 1)(n + 2) (n + 3) …(2n)

Find: Convert into Factorial

Given 1 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 9 …(2n – 1)

Find: Convert into Factorial

=

Given: (2 + 3)! = 2! + 3!

L.H.S = (2 + 3)!

= 5!

R.H.S = 2! + 3!

= (2 × 1) + (3 × 2 × 1)

= 2 + 6

= 8

Hence, L.H.S ≠ R.H.S

Given: (2 × 3)! = 2! × 3!

L.H.S = (2 × 3)!

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

R.H.S = 2! × 3!

= 2 × 1 × 3 × 2 × 1

= 12

Hence, L.H.S ≠ R.H.S

Given: n! (n + 2) = n! + (n + 1)!

R.H.S. = n! + (n + 1)!

= n! + (n + 1)(n + 1 – 1)!

= n! + (n + 1)n!

= n!(1 + n + 1)

= n! (n + 2) = L.H.S

L.H.S = R.H.S

Hence, Proved.

Given (n + 2)! = 60 [(n – 1)!]

To Find : n

⇒ (n + 2) × (n + 1) × (n) × (n-1)! = 60 [(n – 1)!]

⇒(n + 2) × (n + 1) × (n) = 60

⇒ (n + 2) × (n + 1) × (n) = 5 × 4 × 3

n = 3

Hence, n = 3

Given: (n + 1)! = 90 [(n – 1)!]

To Find : n ?

⇒ (n + 1) × (n) × (n-1)! = 60 [(n – 1)!]

⇒ (n + 1) × (n) = 90

⇒ (n + 1) × (n) = 10 × 9

n = 9

Hence, n = 9

Given: (n + 3)! = 56 [(n + 1)!]

To Find : n

⇒ (n + 3) × (n + 2) × (n + 1)! = 56 [(n + 1)!]

⇒ (n + 3) × (n + 2) = 56

⇒ (n + 3) × (n + 2) = 8 × 7

⇒n + 3 = 8

n = 5

Hence, n = 5

Given:

Find: n

⇒ (2n - 1) = 11

⇒2n = 12

n = 6

Hence, n = 6

Given:

L.H.S

R.H.S

L.H.S = R.H.S

Hence, Proved.

Given:

L.H.S

R.H.S

L.H.S = R.H.S

Hence, Proved.

Given:

L.H.S

= R.H.S

L.H.S = R.H.S

Hence, Proved.

Given: 27 boys and 14 girls

We have to select one boy from 27 boys and one girl from 14 girls.

Therefore, a number of ways to select one boy are ²⁷C₁ and similarly the number of ways to select one girl is ¹⁴C₁.

Hence the number of ways to select 1 boy and 1 girl to represent the class in a function is ¹⁴C₁ײ⁷C_{1 =} 14 × 27 = 378

Given: 10 fountain pens, 12 ball pens, and 5 pencil

We have to select one fountain pen from 10 fountain pens, one ball pen from 12 ball pens and one pencil from 5 pencils.

Therefore, the number of ways to select one fountain pen is ¹⁰C₁ and similarly the number of ways to select one ball pen is ¹²C₁ and number of ways to select one pencil from 5 pencils in ⁵C₁

Hence, the number of ways to select one fountain pen, one ball pen and one pencil from a stationery shop is ¹⁰C₁ × ¹²C₁ × ⁵C_{1 =} 10 × 12 × 5 = 600

we have to go from Goa to Delhi via Bombay.

Given: the number of ways from goa to Mumbai is air and sea

Therefore, the number of ways to go from Goa to Mumbai is ²C₁

Given a number of ways from Mumbai to Delhi are air, rail, and road.

Therefore, the number of ways to go from Mumbai to Delhi is ³C₁

Hence the number of ways to go from Goa to Delhi via Bombay is ²C₁ × ³C_{1 =} 2 × 3 = 6

According to the given question, there can be four types of month possible. Months with 30 or 31 days for normal months and for February it can be 28 or 29 days

Now for the months, it can start from any of one day of the week, so there are 7 possibilities.

Hence, the number of types calendars should it prepare to serve for all the possibilities in future years is 7 × 4 = 28

According to the given question one parcel can be posted in 5 ways, that is in either of the one post offices so ⁵C₁. Similarly, for other parcels also it can be posted in ⁵C₁ ways.

Hence the number of ways the parcels be sent by registered post is ⁵C₁ × ⁵C₁ × ⁵C₁ × ⁵C_{1 =} 5 × 5 × 5 × 5 = 625

Given: A coin is tossed 5 times, so each time the outcome is either heads or tails, which implies two possibilities are possible.

So, total possible outcomes possible are ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ = 2 × 2 × 2 × 2 × 2 = 32

Given: An examinee can answer a question either true or false, which implies two possibilities are possible.

So total possible outcomes possible are ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ × ²C₁ = 2¹⁰

Given: Each ring consists of 10 letters and there are three rings. Only one code is correct which will open the lock.

Therefore one subtracted from total possible outcomes will give the number of unsuccessful attempts.

At a time only one letter can appear in each ring

So total number of possible outcomes are ¹⁰C₁ × ¹⁰C₁ × ¹⁰C₁ = 1000

Hence, the number of unsuccessful attempts to open the lock is 1000 - 1 = 999

Given: Multiple choice question, only one answer is correct of the given options.

So, for the first three questions only one answer is correct out of four, similarly, for the remaining three question, only one answer is correct out of two.

Number of outcomes possible is ⁴C₁ × ⁴C₁ × ⁴C₁ for the first three and ²C_{1 ×}²C₁ × ²C₁ for the remaining three.

Hence, total possible outcomes possible are ⁴C₁ × ⁴C₁ × ⁴C₁ × ²C_{1 ×}²C₁ × ²C₁ = 512

(i) Given there are 5 books of mathematics and 6 books of physics. In order to buy one mathematics book number of ways are ⁵C₁ similarly to buy one physics book number of ways are ⁶C₁

Hence, the number of ways a student buy a Mathematics book and a Physics book is ⁵C₁ × ⁶C₁ = 5 × 6 = 30

(ii) Given there is a total of 11 books, so in order to buy either a Mathematics book or a Physics book it means that only one book out of eleven is bought.

Hence, the number of ways in which a student can either buy either a Mathematics book or a Physics book is ¹¹C₁ = 11

Given: seven flags are available and out of which two are needed to make a signal.

From this, we can say, that we have to select two flags out of seven and arrange these two flags to get one signal.

So, number of ways to select two flags out of seven is ⁷C₂. These flags can be arranged in 2! ways one below the other

Hence total number of signals possible are ⁷C₂ × 2! = 42

Given: A total of three prizes are to be given.

Number of ways to select the winner of the first prize is ¹²C₁

Number of ways to select the winner of the second prize is ¹¹C₁ (11 since one student is already given a prize)

Number of ways to select the winner of the third prize is ¹⁰C₁ (10 since two students are already given a prize)

Hence, total number of ways is ¹²C₁ × ¹¹C₁ × ¹⁰C₁ = 12 × 11 × 10 = 1320

Each AP consists of a unique first term and common difference.

Number of ways to select the first term of a given set is ³C₁ = 3

Number of ways to select a common difference of given set is ⁵C₁ = 5

Hence total number of AP’s possible are ³C₁ × ⁵C₁ = 3 × 5 = 15

Given: A total of three positions are to be given.

Number of ways to select principal is ³⁶C₁

Number of ways to select vice-principal is ³⁵C₁ (35 since one position is already given)

Number of ways to select teacher in charge is ³⁴C₁ (34 since two positions are already given)

Hence, total number of ways is ³⁶C₁ × ³⁵C₁ × ³⁴C₁ = 36 × 35 × 34 = 42840

Given: the three-digit number is required without digit repetition

Assume we have three boxes; the first box can be filled with any one of the nine digits (0 not allowed at first place).

Therefore, possibilities are ⁹C₁, the second box can be filled with any one of the nine digits available possibilities are ⁹C₁, the third box can be filled with any one of the eight digits available possibilities are ⁸C₁.

Hence, number of total outcomes possible are ⁹C₁ × ⁹C₁ × ⁸C₁ = 9 × 9 × 8 = 648

Given: The three-digit number is required

Assume we have three boxes, first box can be filled with any one of the nine digits (zero not allowed at first position) therefore possibilities are ⁹C₁, the second box can be filled with any one of the ten digits available possibilities are ¹⁰C₁, third box can be filled with any one of the ten digits available possibilities are ¹⁰C₁.

Hence number of total outcomes possible are ⁹C₁ × ¹⁰C₁ × ¹⁰C₁ = 9 × 10 × 10 = 900

In odd numbers, the last digit consists of (1, 3, 5, 7, 9)

Assume we have three boxes, first box can be filled with any one of the nine digits (zero not allowed at first position) therefore possibilities are ⁹C₁, the second box can be filled with any one of the ten digits available possibilities are ¹⁰C₁, third box can be filled with any one of the five digits (1,3,5,7,9) ⁵C₁

Hence, number of total outcomes possible are ⁹C₁ × ¹⁰C₁ × ⁵C₁ = 9 × 10 × 5 = 450

(i) Given: Five-digit number is required in which the first digit cannot be zero, and the repetition of digits is not allowed.

Assume five boxes, now the first box can be filled with one of the nine available digits, so the possibility is ⁹C₁

Similarly, the second box can be filled with one of the nine available digits, so the possibility is ⁹C₁

the third box can be filled with one of the eight available digits, so the possibility is ⁸C₁

the fourth box can be filled with one of the seven available digits, so the possibility is ⁷C₁

the fifth box can be filled with one of the six available digits, so the possibility is ⁶C₁

Hence, the number of total possible outcomes is ⁹C₁ × ⁹C₁ × ⁸C₁ × ⁷C₁ × ⁶C₁ = 9 × 9 × 8 × 7 × 6 = 27216

(ii) Given: Five-digit number is required in which the first digit cannot be zero, and the repetition of digits is allowed

assume five boxes, now the first box can be filled with one of the nine available digits, so the possibility is ⁹C₁

Similarly, the second box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁

the third box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁

the fourth box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁

the fifth box can be filled with one of the ten available digits, so the possibility is ¹⁰C₁

Hence, the number of total possible outcomes is ⁹C₁ × ¹⁰C₁ × ¹⁰C₁ × ¹⁰C₁ × ¹⁰C₁ = 9 × 10 × 10 × 10 × 10 = 90000

Given: Four-digit number is required which is greater than 7000

Assume four boxes, now in the first box can either be one of the three numbers 7, 8 or 9, so there are three possibilities which are ³C₁

In the second box, the numbers can be any of the four digits left, so the possibility is ⁴C₁

Similarly, for the third box, the numbers can be any of the three digits left, so the possibility is ³C₁

for the fourth box, the numbers can be any of the two digits left, so the possibility is ²C₁

Hence the total number of possible outcomes is ³C₁ × ⁴C₁ × ³C₁ × ²C₁ = 3 × 4 × 3 × 2 = 72

Given: Four-digit number is required which is greater than 8000

Assume four boxes, now in the first box can either be one of the two numbers 8 or 9, so there are two possibilities which is ²C₁

In the second box, the numbers can be any of the four digits left, so the possibility is ⁴C₁

Similarly, for the third box, the numbers can be any of the three digits left, so the possibility is ³C₁

for the fourth box, the numbers can be any of the two digits left, so the possibility is ²C₁

Hence total number of possible outcomes is ²C₁ × ⁴C₁ × ³C₁ × ²C₁ = 2 × 4 × 3 × 2 = 48

Given: Six persons are to be arranged in a row

Assume six seats, now in the first seat, any one of six members can be seated, so the total number of possibilities is ⁶C₁

Similarly, in the second seat, any one of five members can be seated, so the total number of possibilities is ⁵C₁

In the third seat, any one of four members can be seated, so the total number of possibilities is ⁴C₁

In the fourth seat, any one of three members can be seated, so the total number of possibilities is ³C₁

In the fifth seat, any one of two members can be seated, so the total number of possibilities is ²C₁

In the sixth seat, only one remaining person can be seated, so the total number of possibilities is ¹C₁

Hence the total number of possible outcomes = ⁶C₁ × ⁵C₁ × ⁴C₁ × ³C₁ × ²C₁ × ¹C₁ = 6 × 5 × 4 × 3 × 2 × 1 = 720

Given: Nine-digit number is required in which the first digit cannot be zero and the repetition of digits is not allowed.

Assume nine boxes, now the first box can be filled with one of the nine available digits, so the possibility is ⁹C₁

Similarly, the second box can be filled with one of the nine available digits, so the possibility is ⁹C₁

the third box can be filled with one of the eight available digits, so the possibility is ⁸C₁

the fourth box can be filled with one of the seven available digits, so the possibility is ⁷C₁

the fifth box can be filled with one of the six available digits, so the possibility is ⁶C₁

the sixth box can be filled with one of the six available digits, so the possibility is ⁵C₁

the seventh box can be filled with one of the six available digits, so the possibility is ⁴C₁

the eighth box can be filled with one of the six available digits, so the possibility is ³C₁

the ninth box can be filled with one of the six available digits, so the possibility is ² C₁

Hence the number of total possible outcomes is ⁹C₁ × ⁹C₁ × ⁸C₁ × ⁷C₁ × ⁶C₁ = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9 (9!)

Given: Odd number less than 1000 is required.

In order to make the number odd, the last digit has to either of (3, 5, 7)

Assume three boxes, in the first either of the three digits (3,5,7) can be placed, so the possibility is ³C₁

Case 1: Middle digit is zero

If the middle digit is zero, number of ways of placing odd numbers on the second box = 2

Hence, the total number of ways = 3 × 2 = 6 ways

Case 2: Middle digit is an odd number

Number of ways of filling middle box = 2

Number of ways of filling third box = 1

Hence, the total number of ways = 3 × 3 = 9 ways

Hence total number of outcomes possible = 6 + 9 = 15 ways

In the question, it is given that we have to find three - digit numbers with distinct digits which means the digits should be nonrepeating and all the digits should be odd means no even digit.

Numbers by which we form the three digit numbers are 1, 3, 5, 7, 9 only the odd ones.

We will use the concept of multiplication because we have three sub jobs and each job is dependent on the other because a number selected on hundred’s place will not appear in ones and tens place.

The number of ways in which we can form three - digit numbers with odd digits is , 5 × 4 × 3 = 60.

We can also do it by 3 × 4 × 5 = 60 , there are total of 5 choices in the first digit on any place then it becomes 4 in the second digit place because some number has been placed in the first digit out of 5 , so out of rest 4 numbers one number is again consumed at second place so at the end 3 numbers are left to fit the last and third place of our 3 digit number.

In the question, we have to find the possible number of 6 digit numbers formed by the numbers 4, 5, 6, 7, 8, 9 when repetition of digits is not allowed.

We will use the concept of multiplication because there are six sub jobs dependent on each other because a number appearing on any one place will not appear in any other place.

The number of ways in which we can form six digit numbers with the help of given numbers is 6 × 5 × 4 × 3 × 2 × 1 = 6! = 720

The numbers occurring on first place from the left have 6 choices and when one number is placed then number occurring on the second place from the left will have 5 choices and so on one fewer choice will be available to every next place till one occurs.

In the question, we have to find the possible number of 6 digit numbers formed by the numbers 0, 1, 3, 5, 7, 9 when repetition of digits is not allowed.

We will use the concept of multiplication because there are six sub jobs dependent on each other because a number appearing on any one place will not appear in any other place.

Since zero cannot be used in the first position from the left because then it will become a 5 digit number , so we have total of 6 numbers to choose from, but we exclude zero , so we have 5 numbers to choose for the first place , and out of 5 only one number gets occupied , so 4 numbers are left but we had only 5 choices out of 6 and zero can come on the second position from the left side , so remaining choices for the second position are 5 , the number of choices will decrease by one as we keep on going right side.

The number of ways in which we can form six digit numbers with the help of given numbers is 5 × 5 × 4 × 3 × 2 × 1 = 5 × 5! = 120 × 5 = 600

The numbers occurring on first place from left have 5 choices because zero cannot come here and when one number is placed then number occurring on second place from left should have 4 choices but now zero is included in our choice because of the second position , so there will be 5 choice instead of 4 , and so on one fewer choice will be available to every next place until one occurs.

In the question, we have to find the possible number of 4 digit numbers greater than 5000 formed by the numbers 0, 1, 2, 5, 9 when repetition of digits is not allowed.

We will use the concept of multiplication because there are four sub jobs dependent on each other because a number appearing on any one place will not appear in any other place.

For the first place from left we have two choices 5 and 9 because only then our number will be greater than 5000 , for the second place we have 4 choices because out of two one is assigned to the first position from left and total choice of numbers are 5 , so 5 - 1 = 4 , the number of choices will decrease by one as we keep on going right side.

The number of ways in which we can form six digit numbers with the help of given numbers is 2 × 4 × 3 × 2 = 2 × 4! = 2 × 24 = 48

In the question, we have to find the possible number of 4 digits and 2 alphabetical place serial numbers formed by the numbers 0 to 9 and with only 6 English alphabets when repetition of digits and alphabets is not allowed.

We will use the concept of multiplication because there are six sub jobs dependent on each other because a number and alphabet appearing on any one place will not appear in any other place.

In first position from left we have 6 alphabet choices, so in the second position, we will have 5 choices because no repetition is allowed. On the first position from left for digits or third position from the left of serial number we will have 10 choices, and for next position, we will have 9 choices, the number of choices will decrease by one as we keep on going right side. Because on serial numbers we can write zero on the left most position of numbers.

The number of ways in which we can form four digit with two alphabets serial numbers with the help of given data is 6 × 5 × 10 × 9 × 8 × 7 = 151200

Note the positions of the two alphabets is fixed that is the first two and then four digits are placed afterward if the positions were not fixed then the answer would vary.

In the question, we have to find the possible number of 3 digit numbers formed by the numbers 0 to 9 when repetition of digits is not allowed.

We will use the concept of multiplication because there are three sub jobs dependent on each other because a number appearing on any one place will not appear in any other place.

In the first position from left we will have ten choices, in the second position we will have nine choices, and in the third position, we will have eight choices because repetition is not allowed and one digit is occupied in each position.

The number of ways in which we can form three - digit numbers with the help of given data is 10 × 9 × 8 = 720

There will be only one correct combination out of these so the incorrect combinations will be 720 - 1 = 719

In the question, we have to find the possible number of 4 digit numbers formed by the numbers 3, 5, 6, 9 when repetition of digits is not allowed.

We will use the concept of multiplication because there are four sub jobs dependent on each other because a number appearing on any one place will not appear in any other place because all the four numbers are to be used as they are in the code.

The first position from the left will have four choices; the second position will have three choices because one number is consumed in the first position , the number of choices will decrease by one as we keep on going right side.

The number of ways in which we can form four-digit numbers with the help of given data is 4 × 3 × 2 × 1 = 24

In the question, we have to find the possible number of ways in which three jobs I, II, III can be distributed among three guys A, B, C, and one job can be assigned to one person only.

We will use the concept of multiplication because there are three sub jobs dependent on each other because a job assigned to one person cannot be assigned to another person.

The job I has three choices to be assigned and when it is assigned to one person , there are two persons left, and Job II can be assigned to only two persons after assigning Job II , Job III has only one choice because there is only one person left.

The number of ways in which we can distribute three jobs between three persons with the help of given data is 3 × 2 × 1 = 6.

We have to find the possible number of four digit numbers which are less than 4321 and are formed with the numbers 1, 2, 3, 4 when repetition of digits is allowed.

We will use the concept of multiplication because there are four sub jobs dependent on each other because the number 4321 cannot be exceeded and we have to look at repetitions carefully.

First, let us consider all the cases in which we do not assign 4 on the thousand’s place which means we will only assign 1, 2, 3 only and repetition in all other places.

The number of ways in which we can assign values to four digit number, but the thousand’s place will not be assigned 4 when repetition is allowed with the help of given data is 3 × 4 × 4 × 4 = 192.

Now we will consider the case when at thousand’s place there is only 4 and nothing else.

So there will be only one choice on thousand’s place that is 4 , there will be two choices on hundred’s place because there will not be 3 and 4 because then our number will become greater than 4321 and 3 will be dealt later by fixing it in hundred’s place like 4 is fixed in this case. The rest places have all the choices.

The number of ways in which we can assign values to the four-digit number and the thousand’s place is only assigned 4 but hundred’s place is not assigned 3 when repetition is allowed with the help of given data is 1 × 2 × 4 × 4 = 32.

Now we will consider the case when thousand’s place is fixed by 4 and hundred’s place is fixed by 3, ten’s place is fixed by 1.

There is one choice each on thousand’s and hundred’s place, but on ten’s place also there is one choice which is 1 , so a number of choices on one's place will be four.

The number of ways in which we can assign values to four-digit number and the thousand’s place is only assigned 4, hundred’s place is only assigned 3, ten’s place is only assigned 1 when repetition is allowed with the help of given data is 1 × 1 × 1 × 4 = 4.

Now we will consider the case when thousand’s place is fixed by 4 and hundred’s place is fixed by 3, ten’s place is fixed by 2.

There is one choice each on thousand’s and hundred’s place, but on ten’s place also there is one choice which is 2 , so a number of choices on one's place will be one.

So the number coming to my mind will be 4321 , which is one number.

Hence total numbers occurring are 192+32+4+1 = 229.

In the question, we have to find the possible number of 6 digit numbers formed by the numbers 0, 1, 3, 5, 7, 9 when repetition of digits is not allowed.

We will use the concept of multiplication because there are six sub jobs dependent on each other because a number appearing on any one place will not appear in any other place.

The first position from left will have five choices because zero cannot be assigned to that position because then our number will become a five digit number instead of six , the second position will also have five choices because when a number is occupied by the first position then four numbers are left but we ignored zero for the first position and for the second position we can use zero, the number of choices will decrease by one as we keep on going right side.

The number of ways in which we can form six digit numbers with the help of given data is 5 × 5 × 4 × 3 × 2 × 1 = 600

Numbers which are divisible by 10 should have zero at their one’s place , so we will fix zero on one’s place, and the rest of the positions will have choices accordingly.

The number of ways in which we can form six digit numbers when zero is fixed in one’s place and which are divisible by 10 is 5 × 4 × 3 × 2 × 1 × 1 = 5! = 120

We have to find the number of total outcomes when three dices are rolled the outcomes can be different and can be same.

We will use the concept of multiplication because there are three sub jobs dependent on each other and one job is performed one after the other.

The choices of outcomes from one dice are six; there are three dices so choices for each dice will be six.

The number of outcomes which are formed when three dices are rolled one after the other are 6 × 6 × 6 = 6³ = 216

We have to find the number of total outcomes when a coin is tossed three times, the outcomes can be different and can be the same.

We will use the concept of multiplication because there are three sub jobs dependent on each other and one job is performed one after the other.

The choices of outcomes from one coin when flipped once are two; the coin is flipped three times, so each time the number of outcomes will be two.

The number of outcomes which are formed when a coin is flipped thrice, one after the other, are 2 × 2 × 2 = 2³ = 8

The number of outcomes which are formed when a coin is flipped four times, one after the other, are 2 × 2 × 2 × 2 = 2⁴ = 16

The number of outcomes which are formed when a coin is flipped five times, one after the other, are 2 × 2 × 2 × 2 × 2 = 2⁵ = 32

The number of outcomes which are formed when a coin is flipped n times, one after the other, are 2 × 2 × 2………….till n times…… × 2 = 2ⁿ

We have to find the possible number of four digit numbers that are formed with the numbers 1, 2, 3, 4, 5 when repetition of digits is allowed.

We will use the concept of multiplication because there are four sub jobs dependent on each other and are performed one after the other.

There is a total of five choices for every position as repetition is allowed and there is no zero in the given choices of numbers.

The number of ways in which we can form four digit numbers when repetition of digits is allowed along with given numbers 5 × 5 × 5 × 5 = 5⁴ = 625

We have to find the possible number of three-digit numbers that are formed with the numbers 0, 1, 3, 5, 7 when repetition of digits is allowed.

We will use the concept of multiplication because there are four sub jobs dependent on each other and are performed one after the other.

There are four choices for hundred’s position because there is zero also which cannot be used in hundred’s place because then our number will become a two digit number instead of a three digit number, there are five choices for the ten’s place because there are a total of five numbers, hundred's placed in which zero was not included but in ten’s place zero is included and in one’s place there are also five choices because repetition is allowed.

The number of ways in which we can form three digit numbers when repetition of digits is allowed along with given numbers 4 × 5 × 5 = 100

We have to find the possible number of numbers that are formed with the numbers 0, 1, 2, 3, 4, 5 which are less than 1000 when repetition of digits is allowed.

We will use the concept of multiplication because there are sub jobs dependent on each other and are performed one after the other.

First we will make three-digit numbers, There are five choices for hundred’s position because there is zero also which cannot be used in hundred’s place because then our number will become a two digit number instead of a three digit number, there are six choices for the ten’s place because there are a total of six numbers, hundred's placed in which zero was not included but in ten’s place zero is included and in one’s place there are also six choices because repetition is allowed.

The number of ways in which we can form three digit numbers when repetition of digits is allowed along with given numbers 5 × 6 × 6 = 180

Secondly we will make two-digit numbers, There are five choices for ten’s position because there is zero also which cannot be used in ten’s place because then our number will become a one digit number instead of a two digit number, there are six choices for the one’s place because there are a total of six numbers, ten place in which zero was not included but in one’s place zero is included.

The number of ways in which we can form two digit numbers when repetition of digits is allowed along with given numbers 5 × 6 = 30

Thirdly we will form single digit natural numbers, which are given in the question they are five in number, 5 because zero cannot be included because we need natural numbers.

Hence the total number of numbers formed which are less than 1000 and are formed by given numbers are 180+30+5 = 215.

We have to find the possible five digit numbers that are formed by the numbers 0 to 9 which starts from 67 which means first two digits are 6 and 7, when repetition of digits is not allowed.

We will use the concept of multiplication because there are sub jobs dependent on each other and are performed one after the other.

The first position from left is occupied by 6 and second position is occupied by 7 , now we have to make choices on the third , fourth and fifth position.

The third position has eight choices because out of ten 6, and 7 numbers have been used which means we are left with eight choices, for the fourth position we have seven choices because one is used in the third position, for the fifth position we have six choices left.

The number of ways in which we can form five digit numbers when two digits are fixed, and the repetition of digits is not allowed, along with given numbers 1 × 1 × 8 × 7 × 6 = 336.

We have to find the possible number of ways in which we can distribute eight toys among five students when repetition of distribution of toys is allowed.

We will use the concept of multiplication because there are eight sub jobs dependent on each other and are performed one after the other.

The thing that is distributed is considered to have choices, not the things to which we have to give them; it means that in this problem the toys have choices more precisely five choices are there for each toy and children won’t choose any because toys have the right to choose.

The number of ways in which we can distribute toys among five children where repetition of distribution is allowed 5 × 5 × 5 × 5 × 5 × 5 × 5 × 5 = 5⁸ = 390625

We have to find the possible number of ways in which we can post five letters among seven letter boxes when repetition of distribution of letters is allowed.

We will use the concept of multiplication because there are five sub jobs dependent on each other and are performed one after the other.

The thing that is distributed is considered to have choices, not the things to which we have to give them; it means that in this problem the letters have choices more precisely seven choices are there for each letter and letter boxes won’t choose any because letters have the right to choose.

The number of ways in which we can post letters among seven letter boxes where repetition of distribution is allowed 7 × 7 × 7 × 7 × 7 = 7⁵ = 16807.

We have to find the number of total outcomes when three dices are rolled the outcomes can be different and can be the same, but one outcome of at least one dice is made fixed.

We will use the concept of multiplication because there are three sub jobs dependent on each other and one job is performed one after the other.

The choices of outcomes from one dice are six, there are three dices so choices for each dice will be six but the outcome of at least one dice is made fixed which means at most all dices would have a fixed outcome which is 5.

The number of outcomes which are formed when three dices are rolled one after the other but at least one dice is fixed at 5

= total number of possible outcomes – the number of possible outcomes in which 5 does not occur on any dice.

= 6 × 6 × 6 - 5 × 5 × 5 = 216 – 125 = 91

A number of possible outcomes in which 5 does not occur on any dice means that we have deducted 5 from all the six possible outcomes and now only five possible outcomes are there, which will give 125 total outcomes.

We have to find the possible number of ways in which we can put twenty balls in five boxes so that the first box contains only one ball when repetition of distribution of balls is allowed.

We will use the concept of multiplication because there are twenty sub jobs dependent on each other and are performed one after the other.

The thing that is distributed is considered to have choices not the things to which we have to give them, it means that in this problem the balls have choices more precisely four choices are there for each ball and boxes won’t choose any because letters have the right to choose. But one box has the right to choose any one of the twenty balls, so the first box has twenty choices.

The number of ways in which we can put nineteen balls in four boxes where repetition of distribution is allowed

4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 × 4 = 4¹⁹

Hence the total number of ways in which we can put twenty balls in five boxes such that first box contains only one ball = 20 × 4¹⁹.

ILLUSTRATION –

To illustrate suppose there are five balls a, b, c, d, e to be put in three boxes A, B, C such that first box contains only one ball.

Let the first box be A and suppose we have combination of a, A , so the rest combinations would be 2 × 2 × 2 × 2 = 2⁴ , but we have more combinations like b, A ; c, A ; d, A ; e, A these combinations are five , so we add them at the end, and the final answer we get is = 5 × 2⁴.

We have to find the possible number of ways in which we can put five balls in three boxes when repetition of distribution of balls is allowed.

We will use the concept of multiplication because there are five sub jobs dependent on each other and are performed one after the other.

The thing that is distributed is considered to have choices, not the things to which we have to give them; it means that in this problem the balls have choices more precisely three choices are there for each ball and boxes won’t choose any because balls have the right to choose.

The number of ways in which we can put five balls among three boxes where repetition of distribution is allowed 3 × 3 × 3 × 3 × 3 = 3⁵ = 243.

We have to find the possible number of ways in which we can post seven letters among four letter boxes when repetition of distribution of letters is allowed.

We will use the concept of multiplication because there are seven sub jobs dependent on each other and are performed one after the other.

The thing that is distributed is considered to have choices, not the things to which we have to give them; it means that in this problem the letters have choices more precisely four choices are there for each letter and letter boxes won’t choose any because letters have the right to choose.

The number of ways in which we can post letters among four letter boxes where repetition of distribution is allowed 4 × 4 × 4 × 4 × 4 × 4 × 4 = 4⁷.

(i). We have to find the possible number of ways in which we can give four prizes among five students when no boy gets more than one price which means that there is no repetition.

We will use the concept of multiplication because there are four sub jobs dependent on each other and are performed one after the other.

The thing that is distributed is considered to have choices not the things to which we have to give them, it means that in this problem the prizes have choices more precisely first prize will have five choices, second prize will have four choices and the choices will keep on decreasing by one as we go on giving prizes, and students won’t choose any because prizes will have the right to choose.

The number of ways in which we can give four prizes among five students where repetition of distribution is not allowed 5 × 4 × 3 × 2 = 5! = 120.

(ii). We have to find the possible number of ways in which we can give four prizes among five students when any student can get any number of prices which means that there is a repetition of prizes.

We will use the concept of multiplication because there are four sub jobs dependent on each other and are performed one after the other.

The thing that is distributed is considered to have choices, not the things to which we have to give them; it means that in this problem the prizes have choices more precisely five choices are there for each prize and students won’t choose any because prizes have the right to choose.

The number of ways in which we can give four prizes among five students where repetition of distribution is allowed 5 × 5 × 5 × 5 = 5⁴ = 625.

(iii). We have to find the possible number of ways in which we can give four prizes among five students when no student gets all the prizes.

We will use the concept of multiplication because there are four sub jobs dependent on each other and are performed one after the other.

The thing that is distributed is considered to have choices, not the things to which we have to give them; it means that in this problem the prizes have choices more precisely five choices are there for each prize and students won’t choose any because prizes have the right to choose.

The number of ways in which we can give four prizes among five students when no student gets all the prices

= number of ways in which any student can get any number of prices – the number of ways in which one student gets all the prizes.

= 64 – 5 = 59.

There are five students, so any one student will get all the prizes hence there are five choices to give all the prizes to any one student.

We have to find the possible number of ways in which we can illuminate the hall by lighting the lamps which are independent of each other.

We will use the concept of multiplication because there are four sub jobs dependent on each other and are performed one after the other.

We will use a combination technique in this question, which will make us solve the question in an easy way. The combination means selection in rough language.

The combination is denoted by the symbol ^νC_r , which means a number of ways of selecting r objects at a time from n objects.

We will make cases by choosing a specific number of lamps at one time,

Taking one lamp at a time, we can enlighten the hall in ¹⁰C₁ ways.

Taking two lamps at a time, we can enlighten the hall in ¹⁰C₂ ways.

Taking three lamps at a time, we can enlighten the hall in ¹⁰C₃ ways

Taking four lamps at a time, we can enlighten the hall in ¹⁰C₄ ways.

Taking five lamps at a time, we can enlighten the hall in ¹⁰C₅ ways.

Taking six lamps at a time, we can enlighten the hall in ¹⁰C₆ ways.

Taking seven lamps at a time, we can enlighten the hall in ¹⁰C₇ ways.

Taking eight lamps at a time, we can enlighten the hall in ¹⁰C₈ ways.

Taking nine lamps at a time, we can enlighten the hall in¹⁰C₉ ways.

Taking ten lamps at a time, we can enlighten the hall in¹⁰C₁₀ ways.

By using the binomial expansion (1+x)¹⁰ , if we put x = 1, then we will get the sum of these binomial coefficients in which there is extra ¹⁰C₀ term which is equal to 1, so we subtract 1 from the sum.

Therefore the total number of ways of enlightening the hall are = 2¹⁰ – 1

Given: Singles matches are to be played, either a boy plays against a boy, and a girl plays against a girl.

A number of ways to select a boy from team 1 is ⁶C₁. Similarly, Number of ways to select a boy from team 2 is ⁵C₁

Hence number of singles matches between boys is ⁶C₁ × ⁵C₁ = 6 × 5 = 30

A number of ways to select a girl from team 1 is ⁴C₁. Similarly, Number of ways to select a girl from team 2 is ³C₁

Hence number of singles matches between girls is ³C₁ × ⁴C₁ = 4 × 3 = 12

The total number of matches = 30 + 12 = 42.

To find:⁸P₃

⁸P₃ can be written as P(8,3)

We know,

= 8×7×6

= 336

Hence, ⁸P₃ = 336

To find:¹⁰P₄

¹⁰P₄ can be written as P(10,4)

We know,

= 10×9×8×7

= 5040

Hence, ¹⁰P₄ = 5040

To find:⁶P₆

⁶P₆ can be written as P(6,6)

We know,

{∵ 0! = 1}

= 720

Hence, ⁶P₆ = 720

To find: P(6, 4)

We know,

= 6×5×4×3

= 360

Hence, ⁶P₄ = 360

Given: P (5, r) = P (6, r – 1)

To find: value of r

We know,

So, according to question:

P (5, r) = P (6, r – 1)

⇒ (7 – r) (6 – r) = 6

⇒ 42 – 6r – 7r + r² = 6

⇒ 42 – 6 – 13r + r² = 0

⇒ r² – 13r + 36 = 0

⇒ r² – 9r – 4r + 36 = 0

⇒ r(r – 9) – 4(r – 9) = 0

⇒ (r – 9) (r – 4) = 0

⇒ r = 9 or 4

For, P(n, r): r ≤ n

∴ r = 4 {for, P(5, r)}

Given: 5 P(4, n) = 6 P(5, n – 1)

To find: value of n

We know,

So, according to question:

5 P(4, n) = 6 P(5, n – 1)

⇒ (6 – n) (5 – n) = 6

⇒ 30 – 6n – 5n + n² = 6

⇒ 30 – 6 – 11n + n² = 0

⇒ n² – 11n + 24 = 0

⇒ n² – 8n – 3n + 24 = 0

⇒ n(n – 8) – 3(n – 8) = 0

⇒ (n – 8) (n – 3) = 0

⇒ n = 8 or 3

For, P(n, r): r ≤ n

∴ n = 3 {for, P(4, n)}

Given: P(n, 5) = 20 P(n, 3)

To find: value of n

We know,

So, according to question:

P(n, 5) = 20 P(n, 3)

⇒ (n – 3) (n – 4) = 20

⇒ n² – 3n – 4n + 12 = 20

⇒ n² – 7n + 12 – 20 = 0

⇒ n² – 7n – 8 = 0

⇒ n² – 8n + n – 8 = 0

⇒ n(n – 8) – 1(n – 8) = 0

⇒ (n – 8) (n – 1) = 0

⇒ n = 8 or 1

For, P(n, r): n ≥ r

∴ n = 8 {for, P(n, 5)}

Given:ⁿP₄ = 360

To find: value of n

ⁿP₄ can be written as P(n, 4)

We know,

So, according to question:

ⁿP₄ = P(n, 4) = 360

⇒ n (n – 1) (n – 2) (n – 3) = 360

⇒ n (n – 1) (n – 2) (n – 3) = 6×5×4×3

On comparing:

The value of n = 6

Given: P(9, r) = 3024

To find: value of r

We know,

So, according to question:

P(9, r) = 3024

⇒ (9 – r)! = 5!

⇒ 9 – r = 5

⇒ -r = 5 – 9

⇒ -r = -4

∴The value of r = 4

Given: P(11, r) = P (12, r – 1)

To find: value of n

We know,

So, according to question:

P (11, r) = P (12, r – 1)

⇒ (13 – r) (12 – r) = 12

⇒ 156 – 12r – 13r + r² = 12

⇒ 156 – 12 – 25r + r² = 0

⇒ r² – 25r + 144 = 0

⇒ r² – 16r – 9r + 144 = 0

⇒ r(r – 16) – 9(r – 16) = 0

⇒ (r – 9) (r – 16) = 0

⇒ r = 9 or 16

For, P(n, r): r ≤ n

∴ r = 9 {for, P(11, r)}

Given: P(n, 4) = 12. P(n, 2)

To find: value of n

We know,

So, according to question:

P(n, 4) = 12 P(n, 2)

⇒ (n – 2) (n – 3) = 12

⇒ n² – 3n – 2n + 6 = 12

⇒ n² – 5n + 6 – 12 = 0

⇒ n² – 5n – 6 = 0

⇒ n² – 6n + n – 6 = 0

⇒ n(n – 6) – 1(n – 6) = 0

⇒ (n – 6) (n – 1) = 0

⇒ n = 6 or 1

For, P(n, r): n ≥ r

∴ n = 6 {for, P(n, 4)}

Given: P(n – 1, 3) : P(n, 4) = 1 : 9

To find: value of n

We know,

So, according to question:

P(n – 1, 3) : P(n, 4) = 1 : 9

⇒ n = 9

∴The value of n = 9

Given: P(2n – 1, n) : P(2n + 1, n – 1) = 22 : 7

To find: value of n

We know,

So, according to question:

P(2n – 1, n) : P(2n + 1, n – 1) = 22 : 7

⇒ 7(n + 2)(n + 1) = 22×2 (2n + 1)

⇒ 7(n² + n + 2n + 2) = 88n + 44

⇒ 7(n² + 3n + 2) = 88n + 44

⇒ 7n² + 21n + 14 = 88n + 44

⇒ 7n² + 21n – 88n + 14 – 44 = 0

⇒ 7n² – 67n – 30 = 0

⇒ 7n² – 70n + 3n – 30 = 0

⇒ 7n(n – 10) + 3(n – 10) = 0

⇒ (n – 10)(7n + 3) = 0

∴The value of n = 10

Given: P(n, 5) : P(n, 3) = 2 : 1

To find: value of n

We know,

So, according to question:

P(n, 5) : P(n, 3) = 2 : 1

⇒ (n – 3)(n – 4) = 2

⇒ n² – 3n – 4n + 12 = 2

⇒ n² – 7n + 12 – 2 = 0

⇒ n² – 7n + 10 = 0

⇒ n² – 5n – 2n + 10 = 0

⇒ n(n – 5) – 2(n – 5) = 0

⇒ (n – 5) (n – 2) = 0

⇒ n = 5 or 2

For, P(n, r): n ≥ r

∴ n = 5 {∵ P(n, 5)}

To prove: P(1, 1) + 2. P(2, 2) + 3 . P(3, 3) + … + n . P(n, n) = P(n + 1, n + 1) – 1

We know,

Take L.H.S.:

1. P(1, 1) + 2. P(2, 2) + 3. P(3, 3) + … + n . P(n, n)

= 1.1! + 2.2! + 3.3! +………+ n.n!

{∵ P(n, n) = n!}

= (2! – 1!) + (3! – 2!) + (4! – 3!) + ……… + (n! – (n – 1)!) + ((n+1)! – n!)

= 2! – 1! + 3! – 2! + 4! – 3! + ……… + n! – (n – 1)! + (n+1)! – n!

= (n + 1)! – 1!

= (n + 1)! – 1

{∵ P(n, n) = n!}

= P(n+1, n+1) – 1

= R.H.S

Hence Proved

Given: P(15, r – 1) : P(16, r – 2) = 3 : 4

To find: value of r

We know,

So, according to question:

P(15, r – 1) : P(16, r – 2) = 3 : 4

⇒ (18 – r) (17 – r) = 12

⇒ 306 – 18r – 17r + r² = 12

⇒ 306 – 12 – 35r + r² = 0

⇒ r² – 35r + 294 = 0

⇒ r² – 21r – 14r + 294 = 0

⇒ r(r – 21) – 14(r – 21) = 0

⇒ (r – 14) (r – 21) = 0

⇒ r = 14 or 21

For, P(n, r): r ≤ n

∴ r = 14 {for, P(15, r – 1)}

To find: value of n

We know,

So, according to question:

⇒ (n + 5) (n + 4) = 22 (n – 1)

⇒ n² + 4n + 5n + 20 = 22n – 22

⇒ n² + 9n + 20 - 22n + 22 = 0

⇒ n² – 13n + 42 = 0

⇒ n² – 6n – 7n + 42 = 0

⇒ n(n – 6) – 7(n – 6) = 0

⇒ (n – 7) (n – 6) = 0

⇒ n = 7 or 6

∴The value of n can either be 6 or 7

Given: There are five children

To find: The number of ways in which these children can stand in a queue

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴The total number of ways in which five children can stand in a queue

= the number of arrangements of 5 things taken all at a time

= P(5, 5)

{∵ 0! = 1}

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

Hence, Number of ways in which five children can stand in a queue are 120

Given, the total number of teachers in a school = 36

To find: The number of ways in which one principal and one vice-principal can be appointed

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total number of ways in which this can be done

= the number of arrangements of 36 things taken 2 at a time

= P(36, 2)

= 36 × 35

= 1260

Hence, Number of ways in which one principal and one vice-principal are to be appointed out of total 36 teachers in school are 1260

Given: the total number of letters = 4

To find: Number of ordered pairs of letters that can be formed like (E, K) or (S, E) etc.

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total number of ways in which this can be done

= the number of arrangements of 4 things taken 2 at a time

= P(4, 2)

= 4 × 3

= 12

Hence, Number of ways in which ordered pairs of letters, to be used as initials, can be formed from given 4 letters are 12

Given: the total number of books = 4

To find: Number of ways in which these 4 books can be arranged in a shelf

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways

= the number of arrangements of 4 things taken all at a time

= P(4, 4)

{∵ 0! = 1}

= 4 × 3 × 2 × 1

= 24

Hence, Number of ways in which these 4 books can be arranged in a shelf are 24

Given, the total number of letters in ‘NUMBER’ = 6

To find: Number of different 4-letter words, with or without meanings like numb, nume or nmbr etc.

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total number of ways in which this can be done

= the number of arrangements of 6 things taken 4 at a time

= P(6, 4)

= 6 × 5 × 4 × 3

= 360

Hence, Number of different 4-letter words, with or without meanings that can be formed from the letters of the word ‘NUMBER’ are 360

Given: The odd digits are: 1, 3, 5, 7 and 9.

∴ The total number of odd-digits = 5

To find: Total number of three-digit numbers, with distinct digits, with each digit odd like 135 or 159 etc.

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴The total number of ways in which this can be done

= the number of arrangements of 5 things taken 3 at a time

= P(5, 3)

= 5 × 4 × 3

= 60

Hence, total number of three-digit numbers, with distinct digits, with each digit odd are 60

Given: the total number of letters in ‘DELHI’ = 5

To find: Number of words, with or without meanings using all the letters of the word like eldhi or dehil etc.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways in which this can be done

= the number of arrangements of 5 things taken all at a time

= P(5, 5)

{∵ 0! = 1}

= 5 × 4 × 3 × 2 × 1

= 120

Hence, number of words, with or without meanings using all the letters of the word ‘DELHI’ are 120

Given, the total number of letters in ‘TRIANGLE’ = 8

To find: number of words, with or without meanings using all the letters of the word like triangel or angletri etc.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways in which this can be done

= the number of arrangements of 8 things taken all at a time

= P(8, 8)

{∵ 0! = 1}

= 8!

Hence, number of words, with or without meanings using all the letters of the word ‘TRIANGLE’ are 8!

Given: There are two works each of 3 volumes and two works each of 2 volumes

To find: Number of ways in which these books can be arranged in a shelf provided volumes of the same work are not separated

Let w_1, w_2, w_3, w_4, are four works

w₁ has n₁, n₂, n₃ as volumes

w₂ has m₁, m₂, m₃ as volumes

w₃ has a₁, a₂ as volumes

w₄ has b₁, b₂ as volumes

Now, firstly we have to arrange these 4 works like w₂ w₃ w₁ w₄ or w₁ w₂ w₄ w₃

This can be done in 4! ways

Now, we have to separately arrange volumes of these 4 works

w₁ has 3 volumes which can be arranged like n₂ n₁ n₃ or n₃ n₁ n₂

Volumes of w₁ can be arranged in 3! ways

Similarly,

Volumes of w₂ can be arranged in 3! ways

Volumes of w₃ can be arranged in 2! ways

Volumes of w₄ can be arranged in 2! Ways

∴ Total number of ways = 4! × 3! × 3! × 2! × 2!

= 24 × 6 × 6 × 2 × 2

= 3456

Hence, the total number of ways in which these 10 books be placed on a shelf so that the volumes of the same work are not separated are 3456

Given: The number of items in column A = 6 and in column B = 6

A student is asked to match each item in column A with an item in column B

To find: Possible number of correct or incorrect answers which he can give

Let the items of column A are fixed i.e. they arrangement is not changing

Now, we just have to arrange items of column B

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways in which this can be done

= the number of arrangements of 6 things taken all at a time

= P(6, 6)

{∵ 0! = 1}

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Hence, possible number of correct or incorrect answers which a student can give are 720

Given, digits which can be used to make numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9

The number of these digits are 10

To find: Total number of three-digit numbers with no digit repeated

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total number of ways

= the number of arrangements of 10 things taken 3 at a time

= P(10, 3)

= 10 × 9 × 8

= 720

But in these 720 numbers we also included those numbers which are starting from 0 like 023 or 056 etc. Being starting from 0, these are actually two-digit numbers. So, we need to subtract these numbers.

To find these numbers, fix the position of 0 at hundred’s place.

Remaining numbers = 9 (1, 2, 3, 4, 5, 6, 7, 8 or 9)

Arrange these 9 numbers in remaining 2 places.

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total numbers which are starting from 0 are =

= the number of arrangements of 9 things taken 2 at a time

= P(9, 2)

= 9 × 8

= 72

Hence, total number of three-digit numbers with no digit repeated are, 720 – 72 = 648

Given, numbers are 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9

The total number of these digits are 10

To find: 6-digit telephone numbers starting with 35 and no digit appears more than once

So, fix the position of first two digits as 3 and 5

Now, we need to fill these 4 remaining places

Remaining number of digits are 8 (0, 1, 2, 4, 6, 7, 8, 9)

Arrange these 8 numbers at 4 places.

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total number of ways

= the number of arrangements of 8 things taken 4 at a time

= P(8, 4)

= 8 × 7 × 6 × 5

= 1680

Hence, possible number of 6-digit telephone numbers starting with 35 are 1680

Given: Number of boys = 6 and number of girls = 5

To find: Possible number of arrangements in a group photograph

Let boys be b₁, b₂, b₃, b₄, b₅, b₆ and girls be g₁, g₂, g₃, g₄, g₅

Possible arrangements are

b₁ b₂ b₃ b₅ b₆ b₄

g₂ g₄ g₁ g₅ g₃

b₂ b₁ b₅ b₃ b₄ b₆

g₂ g₄ g₅ g₁ g₃

In this arrangement, we are arranging boys and girls separately

Formula used:

Number of arrangements of n things taken all at a point = P(n, n)

∴ Number of ways to arrange boys

= the number of arrangements of 6 things taken all at a time

= P(6, 6)

{∵ 0! = 1}

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Formula used:

Number of arrangements of n things taken all at a point = P(n, n)

∴ Number of ways to arrange girls

= the number of arrangements of 5 things taken all at a time

= P(5, 5)

{∵ 0! = 1}

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

Now, we will get total number of ways by multiplying their separate arrangements

∴ Total number of ways

= 720 × 120

= 86400

Hence, possible number of arrangements in which 6 boys and 5 girls can be arranged for a group photograph with provided conditions are 86400

Given: a = 182 bc

To find: value of x

a denotes the number of permutations of (x + 2) things taken all at a time

∴ a = P(x+2, x+2)

{Number of arrangements of n things taken all at a time = P(n, n)}

b denotes the number of permutations of x things taken 11 at a time

∴ b = P(x, 11)

{Number of arrangements of n things taken r at a time = P(n,r)}

c denotes the number of permutations of x – 11 things taken all at a time

∴ c = P(x-11, x-11)

{Number of arrangements of n things taken all at a time = P(n,n)}

According to question:

a = 182 bc

⇒ P(x+2, x+2) = 182 × P(x, 11) × P(x-11, x-11)

{∵ 0! = 1}

⇒ (x + 2) (x + 1) = 182

⇒ x² + 2x + x + 2 = 182

⇒ x² + 3x + 2 – 182 = 0

⇒ x² + 3x – 180 = 0

⇒ x² – 12x + 15x – 180 = 0

⇒ x(x – 12) + 15(x – 12) = 0

⇒ (x – 12) (x + 15) = 0

⇒ x = -15 or 12

⇒ x = 12 {x cannot hold negative value}

Hence, the value of x is 12

Given: Digits which can be used to make numbers are 1, 2, 3, 4, 5, 6, 7, 8 and 9

The number of these digits are 9

To find: total number of three-digit numbers with no digit repeated

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total number of ways

= the number of arrangements of 9 things taken 3 at a time

= P(9, 3)

= 9 × 8 × 7

= 504

Hence, total number of three-digit numbers using digits 1 to 9 with no digit repeated are 504

Given: Digits which can be used to make numbers are 1, 2, 3, 4, 5, 6 and 7

The number of these digits are 7

To find: total number of three-digit even numbers with no digit repeated

Even numbers are those numbers whose unit’s place is even

∴ fix the position of 1 even number at unit’s place at one time

Even numbers are: 2, 4 and 6

Case 1:

Fix position of 2 at unit’s place

Remaining numbers = 6

Arrange these 6 numbers at remaining 2 places

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total numbers ending with 2 are =

= the number of arrangements of 6 things taken 2 at a time

= P(6, 2)

= 6 × 5

= 30

Case 2:

Fix position of 4 at unit’s place

Remaining numbers = 6

Now, arrange these 6 numbers in remaining 2 places

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total numbers ending with 2 are =

= the number of arrangements of 6 things taken 2 at a time

= P(6, 2)

= 6 × 5

= 30

Similarly,

When you fix position of 6 at unit’s place, 30 more numbers will be formed.

Hence, total number of three-digit even numbers with no digit repeated are, 30 + 30 + 30 = 90

Given, digits which can be used to make numbers are 1, 2, 3, 4 and 5

The number of these digits are 5

To find: total number of 4-digit numbers with no digit repeated

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ The total number of ways

= the number of arrangements of 5 things taken 4 at a time

= P(5, 4)

= 120

Hence, total number of 4-digit numbers using digits 1 to 5 with no digit repeated are 120

Now, for 4-digit even number from digits 1, 2, 3, 4 and 5:

Let 4-digit even number be

Fix the position of unit’s place i.e. t as an even number for which we have 2 choices (2 or 4)

Now, for position x we have remaining 4 choices

Similarly, for position y and z we have 3 and 2 choices respectively

Total number of even numbers are

= multiplication of choices of x y z t

= 4 × 3 × 2 × 2

= 48

Hence, total number of 4-digit numbers using digits 1 to 5 with no digit repeated are 48

Given, word is ‘EAMCOT’

To find: number of arrangements in which no two vowels are adjacent to each other

Vowels in word ‘EAMCOT’ = 3(E, A, O) and consonants = 3(M, C, T)

Let vowels be denoted by V

Now, fix the position by Vowels like this:

The remaining 4 places can be occupied by 3 consonants

Now, arrange 3 consonants at 4 places and 3 vowels at 3 places

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements of vowels

= the number of arrangements of 3 things taken all at a time

= P(3, 3)

{∵ 0! = 1}

= 3!

= 3 × 2 × 1

= 6

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ Total number of arrangements of consonants

= the number of arrangements of 4 things taken 3 at a time

= P(4, 3)

= 4!

= 4 × 3 × 2 × 1

= 24

Hence, total number of arrangements in which no two vowels are adjacent to each other, 6 × 24 = 144

Given: the word is ‘FAILURE.’

To find: number of arrangements so that the consonants occupy only odd positions

Number of vowels in word ‘FAILURE’ = 4(E, A, I, U)

Number of consonants = 3(F, L, R)

Let consonants be denoted by C

Odd positions are 1, 3, 5 or 7

Now, fix the position of consonants like this:

So, arrange these 3 consonants at 4 places

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ Total number of arrangements of consonants

= the number of arrangements of 4 things taken 3 at a time

= P(4, 3)

= 4!

= 4 × 3 × 2 × 1

= 24

Remaining 3 even places and 1 odd place can be occupied by 4 vowels

So, arrange these vowels at remaining places

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements of vowels

= the number of arrangements of 4 things taken all at a time

= P(4, 4)

{∵ 0! = 1}

= 4!

= 4 × 3 × 2 × 1

= 24

Hence, the number of arrangements so that the consonants occupy only odd positions = 24 × 24 = 576

Given: the word is ‘STRANGE.’

To find: a number of arrangements in which vowels come together

Number of vowels in this word = 2(A, E)

Now, consider these two vowels as one entity(AE together as a single letter)

So, the total number of letters = 6(AE S T R N G)

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 6 things taken all at a time

= P(6, 6)

{∵ 0! = 1}

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Two vowels which are together as a letter can be arranged in 2

Ways like EA or AE

Hence, total number of arrangements in which vowels come together = 2 × 720 = 1440

Given: the word is ‘STRANGE.’

To find: a number of arrangements in which vowels never come together

To find out these, we will find the total number of arrangements irrespective of any condition and subtract those arrangements in which vowels come together

Total number of letters in the word = 7

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements irrespective of any condition

= the number of arrangements of 7 things taken all at a time

= P(7, 7)

{∵ 0! = 1}

= 7!

= 7 × 6 × 5 × 4 × 3 × 2 × 1

= 5040

Number of vowels in this word = 2(A, E)

Now, consider these two vowels as one entity(AE together as a single letter)

So, the total number of letters = 6(AE S T R N G)

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 6 things taken all at a time

= P(6, 6)

{∵ 0! = 1}

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Two vowels which are together as a letter can be arranged in 2

Ways like EA or AE

∴ Total number of arrangements in which vowels come together = 2 × 720 = 1440

Hence, the total number of arrangements in which vowel never come together = 5040 – 1440 = 3600

Given: the word is ‘STRANGE.’

To find: number of arrangements so that the vowels occupy only odd positions

Number of vowels in word ‘STRANGE’ = 2(E, A)

Number of consonants = 5(S, T, R, N, G)

Let vowels be denoted by V

Odd positions are 1, 3, 5 or 7

So, fix the position by Vowels like this:

Now, arrange these 2 vowels at 4 odd places

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ Total number of arrangements of vowels

= the number of arrangements of 4 things taken 2 at a time

= P(4, 2)

= 4 × 3

= 12

The remaining 3 even places and 2 odd places can be occupied by 5 consonants

So, arrange these consonants at these places

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements of consonants

= the number of arrangements of 5 things taken all at a time

= P(5, 5)

∵ 0! = 1}

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

Hence, the number of arrangements so that the vowels occupy only odd positions = 12 × 120 = 1440

Given, the word is ‘SUNDAY.’

To find: Number of words that can be formed using letters of the given word and number of words starting with D

Total number of letters in the word = 6

So, arrange these 6 letters

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 6 things taken all at a time

= P(6, 6)

{∵ 0! = 1}

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Hence, the total number of words can be made by letters of the word ‘SUNDAY’ = 720

Now, we need to find out a number of words starting with D

So, fix the position of first letter as D:

Remaining number of letters = 5

Now, arrange these 5 letters at 5 places.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways

= the number of arrangements of 5 things taken all at a time

= P(5, 5)

{∵ 0! = 1}

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

Hence, the possible number of words using letters of ‘SUNDAY’ starting with ‘D’ is 120

Given: the word is ‘ORIENTAL.’

To find: number of arrangements so that the vowels occupy only odd positions

Number of vowels in the word ‘ORIENTAL’ = 4(O, I, E, A)

Number of consonants in given word = 4(R, N, T, L)

Let vowels be denoted by V

Odd positions are 1, 3, 5 or 7

So, fix the position by vowels like this:

Now, arrange these 4 vowels at 4 places

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements of vowels

= the number of arrangements of 4 things taken all at a time

= P(4, 4)

{∵ 0! = 1}

= 4!

= 4 × 3 × 2 × 1

= 24

The remaining 4 even places can be occupied by 4 consonants

So, arrange 4 consonants at remaining places

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements of consonants

= the number of arrangements of 4 things taken all at a time

= P(4, 4)

{∵ 0! = 1}

= 4!

= 4 × 3 × 2 × 1

= 24

Hence, the number of arrangements so that the vowels occupy only odd positions = 24 × 24 = 576

Given: the word is ‘SUNDAY.’

To find: number of words that can be formed with the letters of the given word, that can begin with N, and that can begin with N and end in Y

Total number of letters = 6

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 6 things taken all at a time

= P(6, 6)

{∵ 0! = 1}

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Hence, the total number of words can be made by letters of the word ‘SUNDAY’ = 720

Now, we need to find out of a number of words starting with N

So, fix the position of the first letter as N:

Remaining number of letters in the word ‘SUNDAY’ = 5

Now, we need to arrange these 5 letters at 5 places.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways

= the number of arrangements of 5 things taken all at a time

= P(5, 5)

{∵ 0! = 1}

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

Hence, the possible number of words using letters of ‘SUNDAY’ starting with ‘N’ is 120

Now, we need to find out a number of words starting with N and ending with Y

So, fix the position of first and last letter as N and Y:

Remaining number of letters = 4

Now, we need to arrange these 4 letters at 4 places.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways

= the number of arrangements of 4 things taken all at a time

= P(4, 4)

{∵ 0! = 1}

= 4!

= 4 × 3 × 2 × 1

= 24

Hence, the possible number of words using letters of ‘SUNDAY’ starting with ‘N’ and ending with ‘Y’ are 24

Given: the word is ‘GANESHPURI.’

To find: number of words starting with G

So, fix the position of the first letter as G

Remaining number of letters in the word = 9

Now, we need to arrange these 9 letters at 9 places.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways

= the number of arrangements of 9 things taken all at a time

= P(9, 9)

{∵ 0! = 1}

= 9!

Hence, a possible number of words using letters of ‘GANESHPURI’ starting with ‘G’ is 9!

Given: the word is ‘GANESHPURI.’

To find: number of words starting with P and ending with I

So, fix the position of first and last letter as P and I

Remaining number of letters in the word = 8

Now, we need to arrange these 8 letters at 8 places.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways

= the number of arrangements of 8 things taken all at a time

= P(8, 8)

{∵ 0! = 1}

= 8!

Hence, a possible number of words using letters of ‘GANESHPURI’ starting with ‘P’ and ending with ‘I’ are 8!

Given: the word is ‘GANESHPURI.’

To find: number of words in which vowels are always together

Number of vowels in this word = 4(A, E, I, U)

Now, consider these four vowels as one entity(AEIU together as a single letter) and arrange these letters

So, the total number of letters = 7(AEIU G N S H P R)

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 7 things taken all at a time

= P(7, 7)

{∵ 0! = 1}

= 7!

= 7 × 6 × 5 × 4 × 3 × 2 × 1

= 5040

Now, 4 vowels which are together as a letter can be arranged in 4! (like EAIU or AEUI)

= 4 × 3 × 2 × 1 = 24 ways

∴Total number of arrangements in which vowels come together = 24 × 5040 = 120960

Given: the word is ‘GANESHPURI.’

To find: number of arrangements so that the vowels occupy only even positions

Number of vowels in the word ‘GANESHPURI’ = 4(A, E, I, U)

Number of consonants = 6(G, N, S, H, R, I)

Let a vowel be denoted by V

Even positions are 2, 4, 6, 8 or 10

Now, fix the position by Vowels like this:

Now, arrange 4 vowels at these 5 places

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

Total number of arrangements of vowels

= the number of arrangements of 5 things taken 4 at a time

= P(5, 4)

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

The remaining 1 even place and 5 odd places can be occupied by 6 consonants

So, arrange 6 consonants at these remaining 6 places

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements of consonants

= the number of arrangements of 6 things taken all at a time

= P(6, 6)

{∵ 0! = 1}

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Hence, number of arrangements so that the vowels occupy only even positions = 120 × 720 = 86400

Given: the word is ‘VOWELS.’

To find: number of words that can be formed using letters of the given word

Total number of letters in the word = 6

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 6 things taken all at a time

= P(6, 6)

{∵ 0! = 1}

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Hence, the total number of words can be made by letters of the word ‘VOWELS’ = 720

Given: the word is ‘VOWELS.’

To find: number of words using letters of the given word starting with E

So, fix the position of the first letter as E:

Remaining number of letters = 5

Now, we need to arrange these 5 letters at the remaining 5 places.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways

= the number of arrangements of 5 things taken all at a time

= P(5, 5)

{∵ 0! = 1}

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

Hence, the possible number of words using letters of ‘VOWELS’ starting with ‘E’ is 120

Given: the word is ‘VOWELS.’

To find: number of words using letters of the given word starting with O and ending with L

So, fix the position of first and last letter as O and L:

Remaining number of letters = 4

Now, we need to arrange these 4 letters at the remaining 4 places.

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ The total number of ways

= the number of arrangements of 4 things taken all at a time

= P(4, 4)

{∵ 0! = 1}

= 4!

= 4 × 3 × 2 × 1

= 24

Hence, the possible number of words using letters of ‘VOWELS’starting with ‘O’ and ending with ‘L’ is 24

Given: the word is ‘VOWELS.’

To find: number of words in which vowels always come together

Number of vowels in this word = 2(O, E)

Now, consider these two vowels as one entity(OE together as a single letter)

So, the total number of letters = 5 (OE V W L S)

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 5 things taken all at a time

= P(5, 5)

{∵ 0! = 1}

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

Now, 2 vowels which are together as a letter can be arranged in 2! (like OE or EO)

= 2 × 1 = 2 ways

Total number of words in which vowels come together = 2 × 120 = 240

Given: the word is ‘VOWELS.’

To find: number of words in which consonants always come together

Number of consonants in this word = 4(V, W, L, S)

Now, consider these four consonants as one entity(VWLS together as a single letter)

So, the total number of letters = 3 (VWLS O E)

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 3 things taken all at a time

= P(3, 3)

{∵ 0! = 1}

= 3!

= 3 × 2 × 1

= 6

Now, 4 consonants which are together as a letter can be arranged in 4! (like WLVS or SWLV)

= 4 × 3 × 2 × 1 = 24 ways

Total number of words in which vowels come together = 24 × 6 = 144

Given: the word is ‘ARTICLE.’

To find: number of arrangements so that the vowels occupy only even positions

Number of vowels in word ‘ARTICLE’ = 3(A, E, I)

Number of consonants = 4(R, T, C, L)

Let vowel be denoted by V

Even positions will be 2, 4 or 6

So, fix the position by Vowels like this:

Now, arrange 3 vowels at 3 places

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ Total number of arrangements of vowels

= the number of arrangements of 3 things taken 3 at a time

= P(3, 3)

{∵ 0! = 1}

= 3!

= 3 × 2 × 1

= 6

The remaining 4 odd places can be occupied by 4 consonants

So, arrange 4 consonants at remaining 4 places

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements of consonants

= the number of arrangements of 4 things taken all at a time

= P(4, 4)

{∵ 0! = 1}

= 4!

= 4 × 3 × 2 × 1

= 24

Hence, the number of arrangements so that the vowels occupy only even positions = 6 × 24 = 144

Given: there are 7 married couples

∴ Number of women and men = 7 each

To find: number of ways of selecting 2 men (A, B) and 2 women(C, D) such that C and D are not wives of A and B

Firstly, select 2 men out of a total of 7 men

Formula used:

Number of ways of selecting n things taken r at a time = C(n, r)

∴ Number of ways of selecting 7 things taken 2 at a time

= C(7, 2)

= 7 × 3

= 21

Now, select 2 women out of 5 men(excluding wives of A and B)

Formula used:

Number of ways of selecting n things taken r at a time = C(n, r)

∴ Number of ways of selecting 5 things taken 2 at a time

= C(5, 2)

= 5 × 2

= 10

Numbers of ways in which 4 members of the team can be chosen = 21 × 10

= 210

Now, A can choose B as team partner (⇒ C’s partner is D) or

A can choose C as team partner (⇒ B’s partner is D) or

A can choose D as team partner (⇒ C’s partner is B)

⇒ The team can be formed in these 3 possible ways

∴Number of possible ways in which lawn tennis mixed double be made up from seven married couples if no husband and wife play in the same set = 210 × 3 = 630

Given: there are m men and n women where m>n

To find: out their possible way of sitting arrangements in a row such that no two women sit together

Let m = 2 then possible arrangement is

_ m _ m _

Here 3(2 + 1) gaps for women can by made by 2 men so that no two of them comes together

∴ When m men are there, m seats can be occupied by men and m + 1 seat can by women

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements of men

= the number of arrangements of m things taken all at a time

= P(m, m)

{∵ 0! = 1}

= m!

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ Total number of arrangements of women

= the number of arrangements of m + 1 things taken n at a time

= P(m + 1, n)

Hence, total possible arrangements of m men and n women in a row such that no two women come together

Hence, Proved.

Given: the word is ‘MONDAY.’

Total number of letters in the word = 6

To find: possible number of four-letter words can be made using the letters of the word ‘MONDAY.’

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ Total number of arrangements

= the number of arrangements of 6 things taken 4 at a time

= P(6, 4)

= 6 × 5 × 4 × 3

= 360

Hence, the total number of four-letter words can be made by letters of the word ‘MONDAY’ = 360

Given: the word is ‘MONDAY.’

To find: possible number of words using all the letters of the word ‘MONDAY.’

Total number of letters in the word= 6

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 6 things taken all at a time

= P(6, 6)

{∵ 0! = 1}

= 6!

= 6 × 5 × 4 × 3 × 2 × 1

= 720

Hence, the total number of words can be made by letters of the word ‘MONDAY’ = 720

Given: the word is ‘MONDAY.’

To find: possible number of words using all the letters of the word ‘MONDAY’ but the first letter is a vowel

Total number of vowels = 2(A, O)

Now, fix the position of 1 vowel out of these two at first place which can be done in 2 ways.

or

Now, we need to fill the remaining 5 places

Remaining places for letters = 5

So, arrange 5 letters at 5 places

Formula used:

Number of arrangements of n things taken all at a time = P(n, n)

∴ Total number of arrangements

= the number of arrangements of 5 things taken all at a time

= P(5, 5)

{∵ 0! = 1}

= 5!

= 5 × 4 × 3 × 2 × 1

= 120

Hence, the total number of words can be made by using all letters of the word ‘MONDAY,’ but the first letter is vowel = 120 × 2 = 240

Given: the word is ‘ORIENTAL.’

Total number of letters = 8

To find: possible number of three-letter words can be made using the letters of the word ‘ORIENTAL.’

Formula used:

Number of arrangements of n things taken r at a time = P(n, r)

∴ Total number of arrangements

= the number of arrangements of 8 things taken 3 at a time

= P(8, 3)

= 8 × 7 × 6

= 336

Hence, the total number of three-letter words can be made by letters of the word ‘ORIENTAL’ = 336

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word INDEPENDENCE. It has 12 letters and it has 3 repeated letters ‘N’, ‘D,’ ‘E.’ Of which, the letter N is repeated thrice, the letter D is repeated twice, and the letter E is repeated 4 times in that word. All other letters are distinct.

The problem can now be rephrased as to find total number of permutations of 12 objects (I, N, D, E, P, E, N, D, E, N, C, E) of which three objects are of same type (N, N, N), two objects are of another type (D, D), and four objects are of different type (E, E, E, E).

Total number of such permutations

= 1663200

Hence, a total number of permutations of the word INDEPENDENCE is 1663200.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word INTERMEDIATE. It has 12 letters, and it has 3 repeated letters ‘I,’ ‘T,’ ‘E.’ Of which, the letter I is repeated twice, the letter T is repeated twice, and the letter E is repeated 3 times in that word. All other letters are distinct.

The problem can now be rephrased as to find total number of permutations of 12 objects (I, N, T, E, R, M, E, D, I, A, T, E) of which two objects are of same type (I, I), two objects are of another type (T, T), and three objects are of different type (E, E, E).

Total number of such permutations

= 19958400

Hence, a total number of permutations of the word INTERMEDIATE is 19958400.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word ARRANGE. It has 7 letters, and it has 2 repeated letters ‘A’, ‘R.’ Of which, the letter A is repeated twice, and the letter R is also repeated twice. All other letters are distinct.

The problem can now be rephrased as to find a total number of permutations of 7 objects (A, R, R, A, N, G, E) of which two objects are of same type (A, A), and two objects are of another type (R, R).

Total number of such permutations

= 1260

Hence, a total number of permutations of the word ARRANGE is 1260.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word INDIA. It has 5 letters, and it has 1 repeated letter ‘I.’ The letter I is repeated twice and all other letters are distinct.

The problem can now be rephrased as to find a total number of permutations of 5 objects (I, N, D, I, A) of which two objects are of same type (I, I).

Total number of such permutations

= 60

Hence, a total number of permutations of the word INDIA is 60.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word PAKISTAN. It has 8 letters, and it has 1 repeated letter ‘A.’ The letter A is repeated twice, and all other letters are distinct.

The problem can now be rephrased as to find a total number of permutations of 8 objects (P, A, K, I, S, T, A, N) of which two objects are of same type (A, A).

Total number of such permutations

= 20160

Hence, a total number of permutations of the word PAKISTAN is 20160.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word RUSSIA. It has 6 letters, and it has 1 repeated letter ‘S.’ The letter S is repeated twice, and all other letters are distinct.

The problem can now be rephrased as to find a total number of permutations of 6 objects (R, U, S, S, I, A) of which two objects are of same type (S, S).

Total number of such permutations

= 360

Hence, a total number of permutations of the word RUSSIA is 360.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word SERIES. It has 6 letters, and it has 2 repeated letter ‘S,’ and ‘E.’ The letter S is repeated twice, and letter E is also repeated twice. And all other letters are distinct.

The problem can now be rephrased as to find a total number of permutations of 6 objects (S, E, R, I, E, S) of which two objects are of same type (S, S), and two objects are of another type (E, E).

Total number of such permutations

= 180

Hence, a total number of permutations of the word SERIES is 180.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word EXERCISES. It has 9 letters, and it has 2 repeated letters ‘E’, and ‘S.’ The letter E is repeated thrice, and letter S is repeated twice. And all other letters are distinct.

The problem can now be rephrased as to find total number of permutations of 9 objects (E, X, E, R, C, I, S, E, S) of which three objects are of same type (E, E, E), and two objects are of another type (S, S).

Total number of such permutations

= 30240

Hence, a total number of permutations of the word EXERCISES is 30240.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Given, the word CONSTANTINOPLE. It has 14 letters and it has 3 repeated letters ‘O’, ‘N,’ and ‘T.’ The letter O is repeated twice, the letter N is repeated thrice, and letter T is repeated twice. And all other letters are distinct.

The problem can now be rephrased as to find total number of permutations of 14 objects (C, O, N, S, T, A, N, T, I, N, O, P, L, E) of which two objects are of same type (O, O), three objects are of another type (N, N, N), and the two objects are of different type (T, T).

Total number of such permutations

= 3632428750

Hence, total number of permutations of the word CONSTANTINOPLE is 3632428750.

Given, the word ‘ALGEBRA’. It has 7 letters of which 4 of them are consonants (L, G, B, R) and 3 of them are vowels (A, E, A) repeating the vowel A twice.

We have to find a number of words that can be formed without changing the relative order of vowels and consonants, i.e. if a vowel comes before a consonant in word ALGEBRA, it has to be in the same order in all possible words. For example- ‘ELGABRA’ and ‘AGLEBRA’ are such two words.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of samethe type are in denominator multiplication with factorial.

The consonants in their positions can be arranged in 4! = 24 number of ways, since there is no repeating letter in consonants set (L, G, B, R).

Similarly, the vowels in their positions can be arranged in 3! / 2! = 3 number of ways, since there is one repeated letter in vowel set (A, E, A), i.e. the letter A repeating twice.

Now, total number of arrangements where relative position of consonants and vowels are not changed will be exactly equal to number of ways we can select one to one element from consonant set to vowel set using Multiplication principle (If an event A can occur in m different ways and another event B can occur in n different ways then a total number of ways of simultaneous occurrence of both events in definite order is m x n.)

i.e. Total arrangements = 24 x 3

= 72

Hence, a total number of ways the word ‘ALGEBRA’ be arranged such that the relative position of vowels and consonants are unchanged is equaled to 72.

Given, the word UNIVERSITY. It has 4 vowels and 6 consonants. Vowels included in given word are (U, I, E, I) and consonants included in given word are (N, V, R, S, T, Y). In given word, one vowel letter I is repeated twice.

We have to find a number of words that can be formed in such a way that all vowels must come together. For example- NUIEIVRSTY, UIIENRSYTV, YTSRIIEUVN, and RTSEIUIYNV are few words among them. Notice that all vowel letters come in the bunch (In bold letters).

A specific method is usually used for solving such type of problems. According to that, we assume the group of letters that remain together (here U, I, E, I) are assumed to be a single letter and all other letters are as usual counted as a single letter. Now find a number of ways as usual; the number of ways of arranging r letters from a group of n letters is equals to ⁿP_r. And the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here U, I, E, I).

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Now,

Letters of word UNIVERSITY are U, N, I, V, E, R, S, I, T, Y. (10 letters)

Now from our method, letters are UIEI, N, V, R, S, T, Y. (7 letters)

A total number of arrangements of 7 letters (here all distinct) is 7!

And the total number of arrangements of grouped letters (Here U, I, E, I) is .

So, our final answer for arranging the letters such that all vowels stick together equals multiplication of 7! and

Total number of arrangements

= 60480

Hence, a total number of words formed during the arrangement of letters of word UNIVERSITY such that all vowels remain together is equals to 60480.

Given expression a³b²c⁴ i.e. in expansion aaabbcccc.

To find: Number of expressions that can be generated by permuting the letters of given expression aaabbcccc.

Given expression has three repeating characters a, b, and c. The letter a is repeated 3 times, the letter b is repeated 2 times, and the letter c is repeated 4 times.

So, the given problem can now be rephrased as to find a total number of arrangements of 9 objects (3+2+4) of which 3 objects are of the same type, 2 objects are of another type, and 4 objects are of different type.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

The number of ways of arranging 9 objects of which 3, 2, and 4 objects are of different types is equaled to

= 124

Hence, number of ways of arranging the letters of word/expression aaabbcccc is equals to 124.

Given, the word PARALLEL. It has 8 letters of which 2 letters (A, L) are repeating. The letter A is repeated twice, and the letter L is repeated thrice in the given word PARALLEL.

To find: Number of ways the letters of word PARALLEL be arranged in such a way that not all L’s do come together.

How are we going to find it? First, we find all arrangements of word PARALLEL and then we minus all those arrangements of word PARALLEL in such a way that all L’s do come together, from it. This will exactly be the same as- all number of arrangements such that not all L’s do come together.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

A total number of arrangements of word PARALLEL: Total letters 8. Repeating letters A and L, the letter A repeating twice and letter L repeating thrice. The total number of arrangements

Now we find a total number of arrangements such that all L’s do time together.

A specific method is usually used for solving such type of problems. According to that, we assume the group of letters that remain together (here L, L, L) are assumed to be a single letter, and all other letters are as usual counted as a single letter. Now find a number of ways as usual; the number of ways of arranging r letters from a group of n letters is equals to ⁿP_r. And the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here L, L, L).

Letters in word PARALLEL: 8 letters

Letters in a new word: LLL, P, A, A, R, E: 6 letters (Letter A repeated twice).

Total number of word arranging all the letters

where second fraction 3! divided by 3! comes from arranging letters inside the group LLL: arrangements of three letters where all the three letters are same = 1 (Obviously! You can even think of it).

Now, a Total number of arrangements where not all L’s do come together is equals to total arrangements of word PARALLEL minus the total number of arrangements in such a way that all L’s do come together.

= 3000

Hence, a total number of arrangements of word PARALLEL in such a way that not all L’s do come together equals to 3000.

Given, the word MUMBAI. It has 2 M’s, and total 6 letters.

We have to find a number of words that can be formed in such a way that all M’s must come together. For example- MMBAIU, AIBMMU, UMMABI are few words among them. Notice that all M letters come in the bunch (In bold letters).

A specific method is usually used for solving such type of problems. According to that, we assume the group of letters that remain together (here M, M) are assumed to be a single letter, and all other letters are as usual counted as a single letter. Now find a number of ways as usual; the number of ways of arranging r letters from a group of n letters is equals to ⁿP_r. And the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here M, M).

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Now,

Letters of word MUMBAI are M, U, M, B, A, I (6 letters)

Now from our method, letters are MM, U, B, A, I. (5 letters)

A total number of arrangements of 5 letters (here all distinct) is 5!

And the total number of arrangements of grouped letters (Here M, M) .

So, our final answer for arranging the letters such that all vowels stick together equals multiplication of 5! and

Total number of arrangements

= 5!

= 120

Hence, a total number of words formed during the arrangement of letters of word MUMBAI such that all M’s remains together equals to 120.

Given, digits 1, 2, 3, 4, 3, 2, 1.

To find: Number of numbers formed from given digits such that odd digits always occupies odd positions/places.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

A total number of ways of arranging 3 even digits = 3! / 2! since there is repetition of a digit (2).

A total number of ways of arranging 4 odd digits = 4! / (2! x 2!) since there are twice repetition of digits (1 and 3).

Total number of ways of arranging the digits such that odd digits always occupies odd places =

= 18

Hence, the number of ways of arranging the digits such odd digits always occupies odd places is equals to 18.

Given, the flags: Red, White, and Green. Total of 9 flags and it has 3 repeated flags Green, Red, and White. The flag Green is repeated thrice, and flag Red is repeated 4 times, flag White is repeated twice.

To find: Number of ways of arranging the flags vertically on a flagstaff.

The problem can now be rephrased as to find total number of permutations of 9 objects (R, R, R, R, W, W, G, G, G) of which three objects are of same type (G, G, G), and two objects are of another type (W, W), and 4 objects are of different type (R, R, R, R).

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Total number of such permutations

= 1260

Hence, total number of permutations of the given type is 1260.

Given, the numbers 1, 3, 3, 0. Total of 4 digits, and it has 1 repeated digit 1 repeated twice.

To find: Number of four digit numbers that can be formed using digits 1, 3, 3, 0. Notice that an arrangement in which digit 0 in the first place will not be counted as four digit number. For example- 0331 will not be counted as four digit number since it is a 3 digit number.

The problem can now be rephrased as to find a total number of permutations of 4 objects (1, 3, 3, 0) of which two objects are of same type (1, 1), And all other objects are distinct. But, 0 cannot be in first place (Condition of four digit number).

First, we will find a total number of permutations of these 4 digits and then we will go minus all those permutations in which 0 will come in first place. This will give us exactly number of four-digit numbers that can be formed by permuting the given digits, i.e. 1, 3, 3, 0.

Total number of permutations will be

Number of permutations in which 0 will come at the first place will be equal to (Number of ways we can arrange the remaining digits, i.e. 1, 3, 3 in the remaining three places)

Total number of permutations of given digits forming a four-digit number is equal to

Hence, total number of permutations of 4 digits (1, 3, 3, 0) forming a 4 digit number is 9.

Given, the word ARRANGE. It has 7 letters of which 2 letters (A, R) are repeating. The letter A is repeated twice, and the letter R is also repeated twice in the given word.

To find: Number of ways the letters of word ARRANGE be arranged in such a way that not all R’s do come together.

First, we find all arrangements of word ARRANGE, and then we minus all those arrangements of word ARRANGE in such a way that all R’s do come together, from it. This will exactly be the same as- all number of arrangements such that not all R’s do come together.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

A total number of arrangements of word ARRANGE: Total letters 7. Repeating letters A and R, the letter A repeating twice and letter R repeating twice. The total number of arrangements

Now we find a total number of arrangements such that all R’s do come together.

A specific method is usually used for solving such type of problems. According to that, we assume the group of letters that remain together (here R, R) is assumed to be a single letter and all other letters are as usual counted as a single letter. Now find a number of ways as usual; the number of ways of arranging r letters from a group of n letters is equals to ⁿP_r. And the final answer is then multiplied by a number of ways we can arrange the letters in that group which has to be stuck together in it (Here R, R).

Letters in word ARRANGE: 7 letters

Letters in a new word: A, RR, A, N, G, E: 6 letters (Letter A repeated twice).

Total number of word arranging all the letters

where second fraction 2! divided by 2! comes from arranging letters inside the group RR: arrangements of two letters where all the two letters are same = 2!/2! = 1 (Obviously! You can even think of it).

Now, a Total number of arrangements where not all R’s do come together is equals to total arrangements of word ARRANGE minus the total number of arrangements in such a way that all R’s do come together.

= 900

Hence, the total number of arrangements of word ARRANGE in such a way that not all R’s come together is equals to 900.

Given, the numbers 0, 1, 1, 5, 9. Total of 5 digits, and it has 1 repeated digit 1 repeated twice.

To find: Number of numbers that can be formed using digits 0, 1, 1, 5, 9 in such a way that arranged number is greater than 50000. Notice that an arrangement in which the first digit is either 5 or either 9 will only produce number greater than 50000. We have to find such numbers.

The problem can now be rephrased as to find a total number of permutations of 5 objects (0, 1, 1, 5, 9) of which two objects are of same type (1, 1), And all other objects are distinct. But, either 5 or 9 will be only in the first place (According to question).

First, we will find a total number of permutations of these 5 digits starting with 5 and then we will find the total number of permutations of these 5 digits starting with 9. Addition of these two numbers will give us the required numbers greater than 50000.

Total number of permutations starting with 5 will be equals to permutations of remaining digits (0, 1, 1, 9) in 4 remaining places

Total number of permutations starting with 9 will be equals to permutations of remaining digits (0, 1, 1, 5) in 4 remaining places

Number of permutations in which either 5 or 9 will come at first place will be equal to

= 24

Hence, total number of permutations of 5 digits (0, 1, 1, 5, 9) forming a 5 digit number greater than 50000 is equals to 24.

Given the word SERIES. A total number of letters in it is 6. Two repeating characters S and E, both repeating twice.

To find: Total number of words that can be formed by permuting all digits in such a way that first place and the last will always be occupied by the letter S. For example- SRIEES, SERIES, SREIES are few words among them. Notice the first and last letter of the word is S.

Total number of such words can be formed by permutation of 4 letters (E, R, I, E) in between two S letters; which will be equals to

= 3 x 4

= 12

The denominator factor of 2! is because there is a repeating letter twice (E) in those 4 letters (E, R, I ,E).

Hence, a total number of words permuting the letters of the word SERIES in such a way that the first and last position is always occupied by the letter S is 12.

Given, the word MADHUBANI. It has 9 letters of which 1 letter (A) is repeating twice and all other letters of the word is are distinct.

To find: Number of ways the letters of word MADHUBANI be arranged in such a way that the word must not begin with M but ends with I.

Let's assume that I will be at the end of the word we are forming, and will not be changed its position. We will not take it into consideration now as its position is fixed. Now we have 8 letters to arrange.

First we find all arrangements of word MADHUBANI and then we minus all those arrangements of word MADHUBANI in such a way that the word is starting with the letter M. This will exactly be same as- all number of arrangements such that the word will not begins with the letters M and ends with the letter I (As the letter I was already fixed in the last position).

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

A total number of arrangements of word MADHUBANI excluding I: Total letters 8. Repeating letter A, repeating twice. The total number of arrangements will be equals to

Now we find a total number of arrangements such that the word begins with the letter M.

The total number of arrangements of word MADHUBANI excluding I will be equals to permutation of 7 objects (A, D, H, U, B, A, N) taking all together.

Letters : 7

Repeating letter: A (2 times)

Total number of word arranging all the letters

Now, a Total number of arrangements where the word not starts with M but ends with I (I was already fixed in the last position) will be equals to total arrangements of word MADHUBAN minus the total number of arrangements in such a way that word starts with letter M.

= 17640

Hence, a total number of arrangements of word MADHUBANI in such a way that the word is not starting with M but ends with I equal to 17640.

Given, the numbers 2, 3, 0, 3, 4, 2, 3. Total of 7 digits, and it has 2 repeated digits 2, and 3 repeating twice and thrice respectively.

To find: Number of seven-digit numbers that can be formed using digits 2, 3, 0, 3, 4, 2, 3. Notice that an arrangement in which digit 0 in the first place will not be counted as seven-digit number, i.e. greater than 1 Million (1000000). For example- 0232334 will not be counted as seven-digit number since it is a 6 digit number.

The problem can now be rephrased as to find total number of permutations of 7 objects (2, 3, 0, 3, 4, 2, 3) of which two objects are of same type (2, 2), and three objects are of another type (3, 3, 3), And all other objects are distinct. But, 0 cannot be in first place (Condition of seven digit number).

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

First, we will find a total number of permutations of these 7 digits and then we will go minus all those permutations in which 0 will come in first place. This will give us exactly number of seven-digit numbers that can be formed by permuting the given digits, i.e. 2, 3, 0, 3, 4, 2, 3.

Total number of permutations

Number of permutations in which 0 will come at the first place will be equal to (Number of ways we can arrange the remaining digits, i.e. 2, 3, 3, 4, 2, 3 in the remaining six places)

Total number of permutations of given digits forming a seven-digit number

= 360

Hence, total number of permutations of 7 digits (2, 3, 0, 3, 4, 2, 3) forming a 7 digit number is equals to 360.

Given, that there are 4 different books each having 3 instances, having a total of 12 books.

Lets assume that there are 4 different books B1, B2, B3, and B4. So we have 12 books as -

B1, B1, B1, B2, B2, B2, B3, B3, B3, B4, B4, B4.

To find: Number of arrangements of these 12 books in such a way that all arrangements must be distinct.

The problem can now be rephrased as to find number of permutations of 12 objects in which 3 objects are of one type, 3 objects are of another type, 3 objects are of a third type, and remaining 3 objects are of different type.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permuation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the number of repeated objects of same type are in denominator multiplication with factorial.

The number of permutation of 12 objects with repeating books in factor of 3

= 369600

Hence, total number of permutation of given 12 books will be equals to 369600.

Given the word MATHEMATICS. It has 11 letters of which the letters M, A, and T are repeating and all letters are repeated twice.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Total number of permutations of 11 objects with 3 objects repeating twice

= 4989600

To find the number of words starting with the letter C: This will be equal to permutation of 10 letters (excluded the letter C) where 3 letters (M, A, and T) are repeated twice

= 453600

To find the number of words starting with the letter T: This will be equal to permutation of 10 letters (excluded the letter T) where 2 letters (M, A) are repeated twice, which will be equals to

= 907200

Hence, a total number of words permuting the letters of word MATHEMATICS is 4989600. A total number of words starting with the letter C is 453600. A total number of words starting with the letter T equals 907200.

Given the molecules’ initials A, G, T, and C (All are repeated thrice).

AAAGGGTTTCCC

To find: Number of arrangements of these 12 molecules in such a way that all arrangements must be distinct.

The problem can now be rephrased as to find a number of permutations of 12 objects in which 3 objects are of one type, 3 objects are of another type, 3 objects are of a third type, and remaining 3 objects are of different type.

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

The number of permutation of 12 objects with repeating molecules in the factor of 3

= 369600

Hence, total number of permutation of given 12 molecules will be equals to 369600.

Given, the disc: Red, Yellow, and Green. Total of 9 discs and it has 3 repeated discs Green, Red, and Yellow. The disc Green is repeated twice, and disc Red is repeated 4 times, disc Yellow is repeated thrice.

To find: Number of the arrangement of discs.

The problem can now be rephrased as to find total number of permutations of 9 objects (R, R, R, R, Y, Y, Y, G, G) of which three objects are of same type (Y, Y, Y), and two objects are of another type (G, G), and 4 objects are of different type (R, R, R, R).

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

Total number of such permutations will be

=

= 1260

Hence, total number of permutations of the given type is 1260.

Given, the numbers 1, 2, 0, 2, 4, 2, 4. Total of 7 digits, and it has 2 repeated digits 2, and 4 repeating thrice and twice respectively.

To find: Number of seven-digit numbers that can be formed using digits 1, 2, 0, 2, 4, 2, 4. Notice that an arrangement in which digit 0 in the first place will not be counted as seven-digit number, i.e. greater than 1 Million (1000000). For example- 0122442 will not be counted as seven-digit number since it is a 6 digit number.

The problem can now be rephrased as to find total number of permutations of 7 objects (1, 2, 0, 2, 4, 2, 4) of which two objects are of same type (4, 4), and three objects are of another type (2, 2, 2), And all other objects are distinct. But, 0 cannot be in first place (Condition of seven digit number).

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of the same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

First, we will find a total number of permutations of these 7 digits and then we will go minus all those permutations in which 0 will come in first place. This will give us exactly number of seven-digit numbers that can be formed by permuting the given digits, i.e. 1, 2, 0, 2, 4, 2, 4.

Total number of permutations

Number of permutations in which 0 will come at the first place will be equal to (Number of ways we can arrange the remaining digits, i.e. 1, 2, 2, 4, 2, 4 in the remaining six places)

Total number of permutations of given digits forming a seven-digit number

= 360

Hence, total number of permutations of 7 digits (1, 2, 0, 2, 4, 2, 4) forming a 7 digit number is equals to 360.

Given, the word ASSASSINATION. It has 13 letters of which repeated letters are A, S, I, and N repeating thrice, 4 times, twice and twice respectively.

We have to find number of words that can formed in such a way that all S’s must come together. For example- ASSSSANNIITO, ININAASSSSATO are few words among them. Notice that all S letters comes in bunch (In bold letters).

A specific method is usually used for solving such type of problems. According to that we assume the group of letters that are remains together (here S, S, S, S) are assumed to be a single letter and all other letters are as usual counted as single letter. Now find number of ways as usual; number of ways of arranging r letters from a group of n letters is equals to ⁿP_r. And the final answer is then multiplied by number of ways we can arrange the letters in that group which has to be sticked together in it (Here S, S, S, S).

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the number of repeated objects of same type are in denominator multiplication with factorial.

Now,

Letters of word ASSASSINATION: 13

Now from our method, letters are SSSS, A, A, A, I, I, N, N, T, O. (10 letters)

Total number of arrangements of 10 letters

And total number of arrangements of grouped letters (Here S, S, S, S) is , equals 1.

So, our final answer for arranging the letters such that all vowels sticks together equals multiplication of and

Total number of arrangements

= 151200

Hence, total number of words formed during arrangement of letters of word ASSASSINATION such that all S’s remains together is equals to 151200.

Given, the word INSTITUTE. It has 9 letters and it has 2 repeated letters ‘I’ and ‘T’. The letter I is repeated twice, and letter T is repeated thrice. And all other letters are distinct.

The problem can now be rephrased as to find total number of permutations of 9 objects (I, N, S, T, I, T, U, T, E) of which two objects are of same type (I, I), three objects are of another type (T, T, T).

Since we know, Permutation of n objects taking r at a time is ⁿP_r,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the number of repeated objects of same type are in denominator multiplication with factorial.

Total number of such permutations

= 30240

Hence, total number of permutations of the word INSTITUTE is 30240.

Given the word SURITI.

Arranging the permutations of the letters of the word SURITI in a dictionary:

To find: Rank of word SURITI in dictionary.

First comes, words starting with letter I = 5! = 120

words starting from letter R = 5!/2! = 60

words starting from SI = 4! = 24 (4 letters no repetation)

words starting from SR = 4!/2! = 12 (4 letters, one repetation of I)

words starting from ST = 4!/2! = 12 (4 letters, one repetation of I)

words starting from SUI = 3! = 6 (3 letters no repetation)

words starting from SUR; SURIIT = 1

SURITI = 1

Rank of the word SURITI = 120 + 60 + 24 + 12 + 12 + 6 + 1 + 1

= 236

Hence the rank of the word SURITI in arranging the letters of SURITI in a dictionary among its permutations is 236.

Given the word LATE.

Arranging the permutations of the letters of the word LATE in a dictionary:

To find: Rank of word LATE in dictionary.

First comes, words starting with letter A = 3! (3 letters, no repetation)

words starting with letter E = 3! (3 letters, no repetation)

words starting with L

words starting with LA:

LAET = 1

LATE = 1

Rank of the word LATE = 6 + 6 + 1 + 1

= 14

Hence the rank of the word LATE in arranging the letters of LATE in a dictionary among its permutations is 14.

Given the word MOTHER.

Arranging the permutations of the letters of the word MOTHER in a dictionary:

To find: Rank of word MOTHER in dictionary.

First comes, words starting from letter E = 5! (5 letters no repetation) = 120

words starting from letter H = 5! (5 letters no repetation) = 120

words starting from ME = 4! (4 letters no repetation) = 24

words starting from MH = 4! (4 letters no repetation) = 24

words starting from MOE = 3! (3 letters no repetation) = 6

words starting from MOH = 3! (3 letters no repetation) = 6

words starting from MOR = 3! (3 letters no repetation) = 6

words starting from MOTE = 2! (2 letters no repetation) = 2

words starting from MOTH:

MOTHER = 1

Rank of the word MOTHER = 120 + 120 + 24 + 24 + 6 + 6 + 6 + 2 + 1

= 309

Hence the rank of the word MOTHER in arranging the letters of MOTHER in a dictionary among its permutations is 309.

Given the letters a, b, c, d, and e.

Arranging the permutations of the letters a, b, c, d, and e in a dictionary:

To find: Rank of word debac in dictionary.

First comes, words starting from letter a = 4! = 24

words starting from letter b = 4! = 24

words starting from letter c = 4! = 24

words starting from letter d:

words starting from da = 3! = 6

words starting from db = 3! = 6

words starting from dc = 3! = 6

words starting from dea = 2! = 4

words starting from deb:

debac = 1

Rank of the word debac = 24 + 24 + 24 + 6 + 6 + 6 + 4 + 1

= 95

Hence the rank of the word debc in arranging the letters a, b, c, d, and e in a dictionary among its permutations is 95.

METHOD: 1 (Method of Permutations)

Given 6 ‘+’ and 4 ‘-’ signs to arrange.

To find: Number of arrangements such that no two ‘-’ comes together.

A number of ways of arrangements of 6 ‘+’ signs = 1 (Since all ‘+’ signs are identical).

Arranging all 6 + signs in such a way that now minus sign will only occupy a place in between or edge of + signs.

For example- “P+P+P+P+P+P+P” Arranging 6 + signs gives 7 positions in which a ‘-’ sign can be placed (represented here as P).

Now, number of ways of arrangements of 4 ‘-’ signs in 7 places (Placed all +’s in an alternate manner leading 7 positions) = ⁷P₄

But all 4 ‘-’ signs are identicals hence a number of ways of arrangements of 4 ‘-’ signs among 7 positions = ⁷P₄ / 4!

A total number of arrangments of + and – signs are given by ⁷P₄ / 4!

= 5 x 7

= 35

Hence, a total number of arrangements of + and – signs is 35.

METHOD: 2 (Method of combinations)

All 6 + signs can be arranged in 1 way (All are identical).

Now we have 7 places and 4 ‘-’ signs: So we have to select a position and arrange 4 signs among 7 places. In how many ways can we did it? We can do it in ⁷C₄ number of ways which indeed equal to

= 5 x 7

= 35

Hence, total number of arrangements of + and – signs is 35.

Given the word INTERMEDIATE. IT has 12 words out of which 6 are vowels, and 6 of them are consonants.

(i) the vowels always occupy even places?

i.e., There are 6 vowels, and there can occupy even places, i.e. position number 2, 4, 6, 8, 10, and 12.

A number of ways of arranging 6 vowels among 6 places = 6!/(2! x 3!)

(Since there are 2 repeating vowels I (twice) and E (thrice)).

A number of ways of arranging 6 consonants among 6 places = 6! / 2!

(Since there are 1 repeating consonant T repeating twice).

Total number of ways of arranging vowels and consonants such that vowels can occupy only even positions

= 21600

Hence, a number of ways of arranging INTERMEDIATE’s letters such that vowels can occupy only even positions is equals to 21600.

(ii) The relative order of vowels and consonants does not alter?

A number of ways of arranging vowels = 6! / (2! x 3!)

A number of ways of arranging consonants = 6! / 2!

Total number of ways of arranging the letter of word INTERMEDIATE such that the relative order of vowels and consonants does not alter

= 21600

Hence, a total number of ways of arranging the letters of the word INTERMEDIATE such that the relative orders of vowels and consonants do not change is 21600.

Given the word ZENITH. It has 6 letters.

To find: Total number of words that can be generated by relative arranging the letters of the word ZENITH.

Since it has 6 letters with no repetition, therefore the number of ways of arranging 6 letters on 6 positions is 6! = 720

To find: Rank of word ZENITH when all its permutations are arranged in alphabetical order, i.e. in a dictionary.

First comes, the words starting from the letter E = 5! = 120

words starting from the letter H = 5! = 120

words starting from the letter I = 5! = 120

words starting from the letter N = 5! = 120

words starting from the letter T = 5! = 120

words starting from letter Z:

words starting from ZE:

words starting from ZEH = 3! = 6

words starting from ZEI = 3! = 6

words starting from ZEN:

words starting from ZENH = 2! =2

words starting from ZENI:

words starting from ZENIHT = 1

ZENITH = 1

The rank of word ZENITH = 120 + 120 + 120 + 120 + 120 + 6 + 6 + 2 + 1 + 1

= 616

Hence, the rank of the word ZENITH when arranged in the dictionary is 616.

First letter can be posted in any of the 5 letter boxes i.e. in 5 ways.

Similarly, each one of the other 3 letters can be posted in any of the 5 letter boxes i.e. in 5 ways.

As, the operations are dependent, so, total number of ways

= 5 ⁴.

Ten thousands’ place (10⁴) can be filled with 1 and 2 i.e. ten thousands’ place can be filled in 2 ways. [as, it cannot be filled with 0, because it will become a 4-digit number then]

From the thousands’ place (10³) to the ones’ (10⁰) place can be filled with 0 or 1 or 2, so, 4 places can be filled in 3 ways.

As, the operations are dependent, so, total number of ways

.

The discussion can be shown pictorially as:

1^st woman can draw water from anyone of the 4 taps.

Now, as, it is given, that, no tap should remain unused so, 2^nd woman can draw water from any of the remaining 3 taps.

Similarly, 3^rd women can draw water from any of the remaining 2 taps.

And, 4^th woman have to draw water from the remaining 1 tap.

As, the operations are dependent, so, total number of ways

= 4!

Alternative Approach:

The number of permutations of 4 objects taken 4 at a time is = ⁴P₄ = 4!.

Total possible outcomes in a throw of 3 dice = 6³ = 216.

When, no die shows an even number i.e. all the die shows odd number (1, 3 or 5), then, the number of possible outcomes = 3³ = 27.

Hence, the total number of possible outcomes in a throw of 3 dice in which at least one of the dice shows an even number = 216 - 27

= 189.

As, it is required that two N’s come together, so, let us consider {NN} as a single object.

Thus we have, 5 objects {B}, {A}, {NN}, {A}, {A}, and there are 3 A's.

So, the number of arrangements in which the two N's are together is

= 20

7 men can be arranged in (7-1)! = 6! ways to sit on a round table.

Now, we can place 7 women in 7 empty seats between them so that no two women will be together, and this can be done in 7! ways.

As, the operations are dependent, so, the number of ways in which 7 men and 7 women can sit on a round table such that no two women sit together =6! ×7!

The discussion can be shown pictorially as:

[W = seats between the 7 men(M)]

We have, 9 objects {C}, {O}, {M}, {M}, {I}, {T}, {T}, {E}, {E} and there are 2 M's, 2 T's, 2 E's.

So, the number of words that can be formed out of the letters of the word ‘COMMITTEE’ is

We have, 11 objects {M}, {A}, {T}, {H}, {E}, {M}, {A}, {T}, {I}, {C}, {S} and there are 2 M's, 2 A's, 2 T's.

So, the number of words that can be formed out of the letters of the word ‘MATHEMATICS’ is

6 men can be arranged in (6-1)! = 5! ways to dine at a round table.

Now, if we place 5 women in 6 empty seats between them so that no two women will be together, and this can be done in ⁶P₅ ways i.e. in

As, the operations are dependent, so, the number of ways in which 6 men and 5 women can dine at a round table if no two women sit together.

=5!×6!

The discussion can be shown pictorially as:

[X = 6 empty seats between the 6 men(M)]

5 boys can be seated in 5! ways to sit in a row. Now, we can place 3 girls in 4 empty seats between them so that each girl is between 2 boys, and this can be done in ⁴P₃ = 24 ways.

As, the operations are dependent, so, the number of ways in which 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys

= 2880.

The discussion can be shown pictorially as:

[X = 4 empty seats between 2 boys(B)]

We can see, from 7! onwards the terms are divisible by 14.

∵7!=7×6×5×4×3×2×1

=(7×2)×6×5×4×3

=14×360

So, we should consider up to 6! term to obtain the remainder.

Now,

1! + 2! + 3! + 4! + 5! + 6!

= 1 + 2 + 6 + 24 + 120 + 720

= 873

On dividing 873 by 14, we get 5 as remainder.

∵14×62=868

and 873-868=5

Thousands’ place (10³) can be filled with 1, 2, 3 and 4 i.e. thousands’ place can be filled in 4 ways.

Hundreds’ place (10²) can be filled in 3 ways [i.e. with the remaining 3 digits].

Similarly, tens’ place (10¹) can be filled in 2 ways and ones’ place can be filled only in 1 way.

As, the operations are dependent, so, total number of ways

=24

The discussion can be shown pictorially as:

ALTERNATIVE APPROACH:

The number of permutations of 4 objects taken 4 at a time is = ⁴P₄ = 4! = 24.

Primarily, excluding the 3 things which are to be included, we have to select (r - 3) things from (n - 3), this can be done in ^{n – 3}C_{r – 3} ways.

Now, the 3 particular things can be selected from 3 remaining things only in 1 way.

And the selected r things can be arranged in r! ways.

So, the number of permutations of n different things taking r at a time when 3 particular things are to be included is = r! ^{n – 3}C_{r – 3}

Total number of five-digit telephone numbers = 10⁵ = 100000.

When no digit is repeated, then the number of five-digit telephone numbers = ¹⁰P₅

[as, 5 digits among 10 digits can be arranged in ¹⁰P₅ ways]

=10×9×8×7×6

= 30240

So, the number of five-digit telephone numbers having at least one of their digits repeated is = 100000 – 30240

= 69760.

The discussion can be shown pictorially as: