Permutations

(i) To Find : Value of

Formulae :

•

•

Let,

By using above formula, we can write,

Cancelling (5!) from numerator and denominator we get,

∴x = 504

Conclusion : Hence, value of the expression is 504.

(ii) To Find : Value of

Formula :

Let,

By using the above formula we can write,

Cancelling (29!) from numerator and denominator,

∴x = 32 × 31 × 30

∴x = 29760

Conclusion : Hence, the value of the expression is 29760.

(iii) To Find : Value of

Formula :

Let,

By using the above formula we can write,

Taking (9!) common from numerator,

Cancelling (9!) from numerator and denominator,

∴x = (12 × 11 × 10) - 10

∴x = 1310

Conclusion : Hence, the value of the expression is 1310.

To Prove : LCM {6!, 7!, 8!} = 8!

Formula :

LCM is the smallest possible number that is a multiple of two or more numbers.

Here, we observe that (8!) is the first number which is a multiple of all three given numbers i.e. 6!, 7! and 8!.

Therefore, 8! is the LCM of {6!, 7!, 8!}

Conclusion : Hence proved

To Prove :

Formula : n! = n × (n - 1)!

=

=

= R.H.S.

∴L.H.S. = R.H.S.

Conclusion :

Given Equation :

To Find : Value of x.

Formula :

By given equation,

By using the above formula we can write,

Cancelling (8!) from both the sides,

∴x = 64

Conclusion : Value of x is 64.

(i) Formula :

Let ,

Multiplying and dividing by

From the above formula,

Conclusion :

(ii) Formula :

Let ,

Above equation can be written as

By using above formula,

Conclusion :

Option (i) and (ii) both are false

Proofs :

For option (i),

L.H.S. = (2 + 3)! = (5!) = 120

R.H.S. = (2!) + (3!) = 2 + 6 = 8

∴L.H.S. ≠R.H.S.

For option (ii),

L.H.S. = (2 × 3)! = (6!) = 720

R.H.S = (2!) × (3!) = 4 × 6 = 24

∴L.H.S. ≠R.H.S.

Important Notes : for any two whole numbers a and b,

•

•

Given Equation :

(n + 1)! = 12 × (n - 1)!

To Find : Value of n

Formula :

By given equation,

(n + 1)! = 12 × (n - 1)!

By using above formula we can write,

∴(n + 1) × (n) × (n - 1)! = 12 × (n - 1)!

Cancelling the term (n - 1)! from both the sides,

∴(n + 1) × (n) = 12 …….. eq(1)

∴(n + 1) × (n) = (4) × (3)

Comparing both the sides, we get,

∴n = 3

Conclusion : Value of n is 3.

Note : Instead of taking product of two brackets in eq(1), it is easy to convert the constant term that is 12 into product of two consecutive numbers and then by observing two sides of equation we can get value of n.

Given Equation :

To Find : Value of n

Formula :

By given equation,

(n + 2)! = 2550 × n!

By using above formula we can write,

∴ (n + 2) × (n + 1) × (n!) = 2550 × n!

Cancelling the term (n)! from both the sides,

∴ (n + 2) × (n + 1) = 2550

∴ (n + 2) × (n + 1) = (51) × (50)

Comparing both the sides, we get,

∴n = 49

Conclusion : Value of n is 49.

Note : Instead of taking product of two brackets in eq(1), it is easy to convert the constant term that is 2550 into product of two consecutive numbers and then by observing two sides of equation we can get value of n.

Given Equation :

To Find : Value of n

Formula :

By given equation,

(n + 3)! = 56 × (n + 1)!

By using above formula we can write,

∴(n + 3) × (n + 2) × (n + 1)! = 56 × (n + 1)!

Cancelling the term (n + 1)! from both the sides,

∴ (n + 3) × (n + 2) = 56

∴ (n + 3) × (n + 2) = (8) × (7)

Comparing both the sides, we get,

∴n = 5

Conclusion : Value of n is 5.

Note : Instead of taking product of two brackets in eq(1), it is easy to convert the constant term that is 56 into product of two consecutive numbers and then by observing two sides of equation we can get value of n.

Given Equation :

To Find : Value of n

Formula :

By given equation,

By using above formula,

Cancelling terms (n - 4)! And (2!),

∴ (n - 2) × (n - 3) = 6

∴ (n - 2) × (n - 3) = (3) × (2)

By comparing both the sides,

Conclusion : Value of n is 5.

Note : Instead of taking product of two brackets in eq(1), it is easy to convert the constant term that is 6 into product of two consecutive numbers and then by observing two sides of equation we can get value of n.

Given Equation :

To Find : Value of n

Formula :

By given equation,

By using above formula,

Cancelling terms (n - 2)! , (2!), (2n - 3)! & n, we get,

….. taking 2 common from the term (2n - 2)

∴ (2n - 1) = 11

∴n = 6

Conclusion : Value of n is 6.

Given : n = 15 and r = 12

To Find : Value of at given n and r

Formula :

•

•

Let ,

Substituting n = 15 and r = 12 in above equation,

By using above formula,

Cancelling (12!) from numerator & denominator,

∴x = 455

Conclusion : Value of at n = 15 and r = 12 is 6.

To Prove :

Formula :

L.H.S. = (n + 2) × (n!) + (n + 1)!

= (n + 2) × (n!) + (n + 1) × (n!)

= (n!) × [(n + 2) + (n + 1)]

= (n!) × (2n + 3)

= R.H.S.

∴L.H.S. = R.H.S.

Conclusion : (n + 2) × (n!) + (n + 1)! = (n!) × (2n + 3)

(i) To Prove :

Formula :

Writing (n!) in terms of (r!) by using above formula,

=

Cancelling (r!),

= n(n - 1)(n - 2)…. (r + 1)

= R.H.S.

∴LHS = RHS

Note : In permutation and combination r is always less than n, so we can write n! in terms of r! by using given formula.

(ii) To Prove :

Formula :

by using above formula,

Cancelling (n - r + 1),

= R.H.S.

∴LHS = RHS

(iii) To Prove :

Formula :

by using above formula,

Taking common,

= R.H.S.

∴LHS = RHS

Given: 10 buses running between Delhi and Agra.

To Find: Number of ways a man can go from Delhi to Agra and return by a different bus.

There are 10 buses running between Delhi and Agra so there are 10 different ways to go from Delhi to Agra. The man cannot return from the same bus he went so number of ways are reduced to 9.

These second event occur in completion of first event so there are:10 × 9=90 ways in which a man can go from Delhi to Agra and return by a different bus.

Given: 5 routes from A to B and 3 routes from B to C.

To find: number of different routes from A to C via B.

Let E₁ be the event : 5 routes from A to B

Let E₂ be the event : 3 routes from B to C

Since going from A to C via B is only possible if both the events E₁ and E₂ occur simultaneously.

So there are 5 × 3=15 different routes from A to C via B.

Given: 12 steamers plying between A and B.

To find: number of ways the round trip from A can be made.

i) The steamer which will go from A to B will be returning back,since the given condition is that same steamer should return.

There are 12 steamers available so there are 12 different ways to make around trip between A & B if done on same steamer.

ii)If the return trip is done on different steamer than the once used in trip on going from A to B then the possible number of ways are:12 × 11=132.

(11 because the once used in going from A to B cannot be used in returning hence, reduced by 1.)

To find :number of ways in which 4 people can be seated in a row containing 5 seats.

The possible number of ways in which 4 people be seated in a row containing 5 seats =⁷P₄ (There are 5 places to be filled with 4 persons where arrangement doesn’t matter.)

⁷P₄ …₍ⁿP_r )

=7 × 6 × 5 × 4

=840

To find: number of ways in which 5 ladies draw water from 5 taps.

Condition: no tap remains unused

The condition given is that no well shold remain unused.

So possible number of ways are:5 × 4 × 3 × 2 × 1=120.

Given: three exercises A, B and C consisting of 12, 18 and 10 questions respectively.

To find: number of ways in which three questions be selected choosing one from each exercise.

Ways of selecting one question from exercise A:¹²C₁ (way of selecting one element from n number of elements.)

Ways of selecting one question from exercise B:¹⁸C₁

Ways of selecting one question from exercise C:¹⁰C₁

So number of ways of choosing one question from each exercise A ,B,C =¹²C₁ × ¹⁸C₁ × ¹⁰C₁

=12 × 18 × 10

=2160

Given: there are four sections of 40 students each in XI standard.

To find : number of ways in which a set of 4 student representatives be chosen, one from each section.

Ways of selecting one student from section 1:⁴⁰C₁

Ways of selecting one student from section 2:⁴⁰C₁

Ways of selecting one student from section 3:⁴⁰C₁

Ways of selecting one student from section 4:⁴⁰C₁

So number of ways of choosing a set of 4 student representatives one from each section=⁴⁰C₁ × ⁴⁰C₁ × ⁴⁰C₁ × ⁴⁰C₁

=40 × 40 × 40 × 40

=2560000

To find: number of ways in which a vowel, a consonant and a digit be chosen out of the 26 letters of the English alphabet and the 10 digits.

e.g.

Way of selecting a vowel from 5 vowels=⁵C₁

Way of selecting a consonant from 26 consonants=²¹C₁

Way of selecting a digit from 10 digits=¹⁰C₁

So ways of choosing a vowel, a consonant, a digit=⁵C₁ × ²¹C₁ × ¹⁰C₁

=5 × 21 × 10

=1050

Given:8 digit telephone number starts with 270 .

To find: How many 8-digit telephone numbers can be constructed?

There are 10 digits between 0 to 9,and three of them are utilized in filling up the first three digits i.e.270 of the 8 digit phone number,so remaining number of digits=10-3=7,and this need to be used in filling up the remaining 8-3=5 places of the telephone number.

i.e.the remaining 5 places need to filled up with any one of:1,3,4,5,6,8,9

So, number of ways=7 × 6 × 5 × 4 × 3=2520.

a) A coin is tossed three times

So possible number of outcomes=2³=8

(HHH,HHT,HTH,HTT,THH,THT,TTH,TTT)

b)

i) A coin is tossed four times

So possible number of outcomes=2⁴=16

(HHHH,HHHT,HHTH,HHTT,HTHH,HTHT,HTTH,HTTT,THHH,THHT,THTH,THTT,TTHH,TTHT,TTTH,TTTT)

ii) A coin is tossed n times

So possible number of outcomes=2ⁿ

Given:5 Flags

Way of generating signal using 2 different flags =⁵P₂ (way of selecting 2 things out of 5 things with considering arrangement.)

Way of generating signal using 3 different flags =⁵P₃

Way of generating signal using 4 different flags =⁵P₄

Way of generating signal using 5 different flags =⁵P₅

So total number of ways =⁵P₂ + ⁵P₃ + ⁵P₄ + ⁵P₅

=20 + 60 + 120 + 120

=320

Given: first 10 letters of the English alphabet.

In 4 letter code for first position there are 10 possibilties for second position there are 9 possibilities ,for third position there are 8 possibilities and for fourth position there are 7 posiibilities since repetition is not allowed.

So total numbers of combination=10 × 9 × 8 × 7=5040

This is the example of Cartesian product of two sets.

The pairs in which the first entry is an element of A and the second is an element of B are :

(2,0),(2,1),(3,0),(3,1),(5,0),(5,1)

⇒3 × 2=6

Given: Two sets: {1, 2, 3} & {2, 3, 4}

To find: number of A.P. with 10n terms whose first term is in the set {1, 2, 3} and whose common difference is in the set {2, 3, 4}

Number of arithmetic progressions with 10 terms whose first term are in the set {1, 2, 3} and whose common difference is in the set {2, 3, 4} are:3 × 3=9

(3 because there are three elements in the set {1, 2, 3} and another 3 because there are three elements in the set {2, 3, 4})

As we can see that For Item2 there can be any of the match

So, For each item in column A there are 6 different options in column B since we don’t have to think about correct or incorrect matching.

So possible number of combinations possible to answer:

6 × 5 × 4 × 3 × 2 × 1=720

To find: types of February calendars that can be prepared.

There are two factors to develop FEBRUARY metallic calendars

1)The day on the start of the year of which possibility=7

2)whether the year is leap year or not of which possibility is =2

So, number of FEBRUARY calendars possibilities to serve in future years=7 × 2=14

Given: 36 teachers are there in a school.

To find: Number of ways in which one principal and one vice-principal can be appointed.

There are 36 options of appointing principal and 35 option of appointing vice-principal since same teacher cannot be appointed as principal and vice-principal.

Total number of ways=36 × 35=1260

A bulb can be good or defective ,so there are 2 different possibilities of a bulb.

So number of all possibile outcomes(of all bulbs)=2 × 2 × 2=8

Given: a set of five true – false questions.

To find: the maximum number of students in the class.

Condition: no student has written the all correct answer and no two students have given the same sequence of answers.

The total number of answering a set of 5 true or false question=2⁵=32

Since, no two students have given the same sequence of answers and no student has written the all correct answer.

Therefore total possibilities reduces by 1(of no student has written the all correct answer)

⇒2⁵-1=32-1=31

Given: 20 students.

The number of ways of giving first and second prizes in mathematics to a class of 20 students=20 × 19.

(First prize can be given to any one of the 20 students but the second prize cannot be given to the student that received the first prize so the number of candidates for the second prize is 19.)

The number of ways of giving first and second prizes in chemistry

to a class of 20 students=20 × 19.

The number of ways of giving first prize in physics to a class of 20 students=20

The number of ways of giving first prize in english to a class of 20 students=20

So total numbe of ways=20 × 19 × 20 × 19 × 20 × 20=57760000

Given: 5 objective-type question, each question having 4 choices.

To find: the number of ways of answering them.

Each objective-type question has 4 choices.

So the total number of ways of answering 5 objective-type question, each question having 4 choices=4 × 4 × 4 × 4 × 4=4⁵

Given: A gentleman has 6 friends to invite. He has 3 servants to carry the cards.

Each friend can be invited by 3 possible number of servants.

So the number of ways of inviting 6 friends using 3 servants=3 × 3 × 3 × 3 × 3 × 3=3⁶

Given:6 rings and 4 fingers.

Each ring has 4 different fingers that they can be worn.

So total number of ways in which 6 rings of different types can be worn in 4 fingers =4 × 4 × 4 × 4 × 4 × 4=4⁶

Each letter has 4 possible letter boxes option.

So the number of ways in which 5 letters can be posted in 4 letter boxes =4 × 4 × 4 × 4 × 4=4⁵ (Each 4 for each letter.)

i)if repetition of letters is not allowed then number of many 3-letters words that can be formed using a, b, c, d, e are

5 × 4 × 3=60

ii) if repetition of letters is allowed then number of many 3-letters words that can be formed using a, b, c, d, e are

5 × 5 × 5=125

To find: Number of 4 digit numbers when a digit may be repeated any number of times

The first place has possibilities of any of 9 digits.

(0 not included because 0 in starting would make the number a 3 digit number.)

The second place has possibilities of any of 10 digits.

The third place has possibilities of any of 10 digits.

The fourth place has possibilities of any of 10 digits.

Since repetition is allowed.

So there are 9 × 10 × 10 × 10=9000 4-digit numbers when a digit may be repeated any number of times.

To find: number of numbers that can be formed from the digits 1, 3, 5, 9 if repetition of digits is not allowed

Forming a 4 digit number:4!

Forming a 3 digit number:⁴C₃ × 3!

Forming a 2 digit number:⁴C₂ × 2!

Forming a 1 digit number:4

So total number of ways=4! + ( ⁴C₃ × 3!) + ( ⁴C₂ × 2!) + 4

=24 + 24 + 12 + 4

=64

In forming a 3 digit number the 100’s place can be occupied by any 9 out of 10 digits(0 not included because it will lead to formation of 2 digit number.)

The 10’s place can be occupied by any of the remaining 9 digits(here 0 can or cannot be used.)

In one’s place any of the remaing 8 digits can be used.

So total 3-digit numbers with no digit repeated are:9 × 9 × 8=648.

100’s place 10’s place Unit’s place

There are total 5 digits available ,for forming a 3 digit number,in 100’s place only 1,3,5,7 can be used(0 not included because it will lead to formation of 2 digit number.)

In 10’s place any of the 5 can be used and same is the case with one’s place.

So total number of 3 digit numbers formed=4 × 5 × 5=100

There are total 6 digits available ,for forming a 6 digit number,in 100000’s place only 1,3,5,7,9 can be used(0 not included because it will lead to formation of 2 digit number.)

In 10000’s place any of the remaining 5 digits can be used(even 0 can be used.)

In 1000’s place any of the remaining 4 digits can be used.

In 100’s place any of the remaining 3 digits can be used.

In 10’s place any of the remaining 2 digits can be used.

In one’s place the remaining digit can be used.

So total number of 6 digit numbers possible=5 × 5 × 4 × 3 × 2 × 1=600

For finding the number of 6 digit numbers divisible by 10 the one’s place should contain 0 so possibilities=5 × 4 × 3 × 2 × 1=120

To find: number of natural numbers less than 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5 when a digit may be repeated any number of times

For forming a 3 digit number less than 1000 possible ways are:

5 × 6 × 6…(in 100’s place 5 digits are only possible 0 not included.)

=180

For forming a 2 digit number less than 1000 possible ways are:

5 × 6……(in 10’s place 5 digits are only possible 0 not included.)

=30

For forming a 1 digit number less than 1000 possible ways are:

5…(0 not included because it is a whole number and natural number is asked in question.)

So total number of numbers less than 1000 that can be formed from the digits 0, 1, 2, 3, 4, 5 when a digit may be repeated any number of times=180 + 30 + 5=215

To find: 6-digit telephone numbers that can be constructed using the digits 0 to 9.

Condition: each number starts with 67 and no digit appears more than once

There are 10 digits between 0 to 9,and two of them are utilized in filling up the first two digits i.e.67 of the 6 digit phone number, so remaining number of digits=10-2=8,and this need to be used in filling up the remaining 6-2=4 places of the telephone number.

So, number of ways=8 × 7 × 6 × 5=1680.

Given: three jobs, I, II and III to be assigned to three persons A, B and C.

To find: In how many ways this can be done.

Condition: one person is assigned only one job and all are capable of doing each job.

It is given that one person is assigned only one job and all are capable of doing each job.

So if for person one 3 options are available ,for peson two 2 options and for person three only one option is available.

So total number of ways in which three jobs, I, II and III be assigned to three persons A, B and C if one person is assigned only one job and all are capable of doing each job=3 × 2 × 1=6

The number of sequences possible =10 × 9 × 8=720(since no repeated digits is the given condition.)

There will be only one successful attempt so the number of unsuccessful attempts to open the lock=720-1=719.

Given: code consists of digits 3, 5, 6, 9.

To find: the largest possible number of trials necessary to obtain the correct code.

The customer remembers that this 4 digit code consists of digits 3, 5, 6, 9.

So the largest possible number of trials necessary to obtain the correct code=4!=4 × 3 × 2 × 1=24

i)To distribute 3 prizes among 4 girls where no girl gets more than one prize the possible number of permutation possible are:⁴P₃=24

ii) To distribute 3 prizes among 4 girls where a girl may get any number of prizes the number of possibilities are:4 × 4 × 4=64.

(since a prize can be given to any of the 4 girls.)

iii) To distribute 3 prizes among 4 girls where no girl gets all the prizes the number of possibilities are:(4 × 4 × 4)-(4)=64-4=60

(the situation where a single girl gets all the prizes has to reduced from the situation where a girl may get any number of prizes.)

To find: the value of ¹⁰P₄

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

Therefore,

¹⁰P₄

¹⁰P₄10 × 9 × 8 × 7

¹⁰P₄5040

Thus, the value of ¹⁰P₄ is 5040.

To find: the value of ⁶²P₃

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

Therefore,

⁶²P₃

⁶²P₃ = 62 × 61 × 60 × 59 = 226920

Thus, the value of ⁶²P₃ is 226920.

To find: the value of ⁶P₆

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

Therefore,

⁶P₆

⁶²P₃ = 6 × 5 × 4 × 3 × 2× 1= 720

Thus, the value of ⁶P₆ is 720.

To find: the value of ⁹P₀

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

Therefore,

⁹P₀

⁹P₀ = 1

Thus, the value of ⁹P₀ is 1.

To Prove: ⁹P₃ + 3 × ⁹P₂ = ¹⁰P₃

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

The equation given below needs to be proved i.e

⁹P₃ + 3 × ⁹P₂ = ¹⁰P₃.

+

(9 × 8 × 7) + (3 × 9 × 8) = 10 × 9 × 8

10 × 9 × 8 = 10 × 9 × 8

Hence, proved.

⁹P₃ + 3 × ⁹P₂ = ¹⁰P₃.

(i) To find: the value of n

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

ⁿP₅ = 20 × ⁿP₃.

20 = (n - 3)(n - 4)

(n - 8)(n + 1) = 0

n = 8, -1

We know, that n cannot be a negative number.

Hence, value of n is 8

(ii) To find: the value of n

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

16 ×ⁿP₃ = 13 × ⁿ⁺¹P₃.

16n – 32 = 13n + 13

3n = 45

n = 15

Hence, value of n is 15.

(iii) To find: the value of n

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

²ⁿP₃ = 100ⁿP₂

2n(2n - 1)(2n - 2) = 100 × n(n - 1)

4n(2n - 1)(n - 1) = 100 × n(n-1)

8n(n - 13) = 0

n = 0, 13

We know that n should be greater than zero.

Hence, value of n is 13

(i) To find: the value of r

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

⁵P_r = 2⁶P_r-1

r = 10, 3

Hence, value of r is 3, 10

(ii) To find: the value of r

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

²⁰P_r = 13²⁰P_r-1

21 – r = 13

r = 8

Hence, value of r is 8.

(iii) To find: the value of r

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

¹¹P_r = ¹²P_r-1

(r - 16)(r - 9) = 0

r = 16, 9

Since r cannot be 16 as it creates a negative factorial in denominator. Therefore,
r = 16 is not possible.

Hence, value of r is 9.

To find: the value of n

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

ⁿP₄ : ⁿP₅ = 1:2

n – 4 = 2

n = 6

Hence, value of n is 6.

(ii) To find: the value of n

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

^n–1P₃ : ⁿ⁺¹P₃, = 5 : 12

12

n = 8, 2.25

Since n cannot be 2.25 as it creates a negative factorial in denominator. Therefore,
n = 2.25 is not possible.

Hence, value of n is 8.

To find: the value of r

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

¹⁵P_r–1 : ¹⁶P_r–2, = 3 : 4

Since r cannot be 21 as it creates a negative factorial in denominator. Therefore,
is not possible.

Hence, value of r is 14

To find: the value of n

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

^2n–1P_n : ²ⁿ⁺¹P_n–1, = 22 : 7

n = 10, -0.42

Since n cannot be -0.42

Hence, value of n is 10.

To find: the value of n

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

ⁿ⁺⁵P_n+1 = (n – 1). ⁿ⁺³P_n

(n - 7)(n - 6) = 0

n = 7, 6

Hence, values of n are 7 & 6

To Prove: 1 + 1. ¹P₁ + 2. ²P₂ + 3. ³P₃ + …. n. ⁿP_n = ⁿ⁺¹P_n+1.

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

1 + 1. ¹P₁ + 2. ²P₂ + 3. ³P₃ + …. n. ⁿP_n = ⁿ⁺¹P_n+1.

1 + (2! - 1!) + (3! - 2!) + (4! - 3!) + …….((n + 1)! - n!) = (n + 1)!

1 + ((n + 1)! - 1!) = (n + 1)!

(n + 1)! = (n + 1)!

Hence proved.

To find: the number of permutations of 10 objects, taken 4 at a time.

Formula Used:

Total number of ways in which n objects can be arranged in r places (Such that no object is replaced) is given by,

ⁿP_r

¹⁰P₄

¹⁰P₄10

¹⁰P₄5040

Hence, the number of permutations of 10 objects, taken 4 at a time is 5040.

To find: number of arrangements of 5 people in 3 seats.

Consider three seats ABC

Now, place A can be occupied by any 1 person out of 5.

Then place B can be occupied by any 1 person from remaining 4 and for C there are 3 people to occupy the seat.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, permutation of 5 different objects in 3 places is

P(5,3) =

= = = 60.

Therefore, the number of possible solutions is 60.

To find: number of arrangements of 7 people in a queue.

Here there are 7 spaces to be occupied by 7 people.

Therefore 7 people can occupy first place.

Similarly, 6 people can occupy second place and so on.

Lastly, there will be a single person to occupy the 7 positions.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, permutation of 7 different objects in 7 places is

P(7,7) =

= = = 5040.

Therefore, the number of possible ways is 5040

To find: number of arrangements of 5 children in a queue.

Here, 5 places are needed to be occupied by 5 children.

Therefore any one of the 5 children can occupy first place.

Similarly, any 4 children can occupy second place and so on.

Lastly, there will be a single person to occupy the 5 position

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, permutation of 5 different objects in 5 places is

P(5,5) =

= = = 120.

Hence, this can be done in 120 ways.

To find: number of arrangements of 6 women drawing water from 6 wells

Here, 6 wells are needed to be used by 6 women.

Therefore any one of the 6 women can draw water from the 1 well.

Similarly, any 5 women can draw water from the 2^nd well and so on.

Lastly, there will be single women left to draw water from the 6^th well.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, permutation of 6 different objects in 6 places is

P(6,6) =

= = = 720.

Hence, this can be done in 720 ways.

To find: number of arrangements of 4 different books in a shelf.

There are 4 different books.

Any one of the four different books can be placed on the shelf first.

Similarly, in the next position, 1 book out of 3 can be placed.

Finally, the last book will have a single place to fit.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, permutation of 4 different objects in 4 places is

P(4,4) =

= = = 24.

Hence they can be arranged in 24 ways.

To find: number of arrangements of names on a ballot paper.

There are six contestants contesting in the elections.

Name of any 1 student out of six can appear first on the ballot paper.

2 position on the ballot paper can be filled by rest of the five names and so on.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, permutation of 6 different objects in 6 places is

P(6,6) =

= = = 720.

Hence, their name can be arranged in 720 ways.

To find: number of arrangements in which women sit in even places

Condition: women occupy even places

Here the total number of people is 8.

In this question first, the arrangement of women is required.

The positions where women can be made to sit is 2^nd, 4^th, 6^th, 8^th. There are 4 even places in which 3 women are to be arranged.

Women can be placed in P (4,3) ways. The rest 5 men can be arranged in 5! ways.

Therefore, the total number of arrangements is P (4,3) ×5!

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 4 different objects in 3 places and the arrangement of 5 men are

P (4,3) ×5! =

=

= 2880.

Hence number of ways in which they can be seated is 2880.

To find: number of possibilities of a selection of answers.

Each item in column A can select another item in column B.

Therefore the question involves selecting each item from column A to each item in column B. this can be done in P(6,6)

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 6 different objects in 6 places is

P(6,6) =

= = = 720.

Therefore, the possible number of selecting an incorrect or correct answer is 720.

(i) the number of initials is 1

In this case, all letters have one chance (i.e. letters F, K, R, V ).

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 4 different objects in 1 place is

P(4,1) =

= = = 4.

So no of ways is 4 .

(ii) the number of initials is 2

There are two cases here

(a) When two R do not occur in initials

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 4 different objects in 2 places is

P(4,2) =

= = = 12.

A number of arrangements here are 12.

(b) When two R occurs in initials

When two R are chosen then 1 pair is included twice.

Selection of 0 letters remaining from 3 letters can be done in P(3,0) ways.

Formula:

A number of permutations of n objects in which p objects are alike of one kind are =n!/p!

Selections = P(3,0) ×

= × = 1

Therefore, the total number of pairs 13.

(iii) the number of initial is 3

(a) two R do not occur in initials

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 4 different objects in 3 places is

P(4,3) =

= = = 24.

A number of arrangements here are 24.

(b) two R occurs in initials

When two R are chosen then 1 pair is included twice.

Selection of 1 letter from the remaining 3 letters is P(3,1)

Formula:

A number of permutations of n objects in which p objects are alike of one kind = n!/p!

Selections = P(3,1) ×

= × = 9

total number of arrangements for 3 initials are 33

(iv) The number of initials is 4

(a) Two R do not occur in initials

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 4 different objects in 4 places is

P(4,4) =

= = = 24.

A number of arrangements here are 24.

(b) Two R occurs in the initials

When two R are chosen then 1 pair is included twice.

Selection of 2 letters from the remaining 3 letters is P(3,2)

Formula:

A number of permutations of n objects in which p objects are alike of one kind = n!/p!

Selections = P(3,2) ×

= × = 36

total number of arrangements for 4 initials are 60

(v) The number of initials is 5

Formula:

A number of permutations of n objects in which p objects are alike of one kind = n!/p!

Selections = = 60.

Total number of arrangements are 4 + 13 + 33 + 60 + 60 = 170

To find: number of ways of winning the first three prizes.

The first price can go to any of the 10 students.

The second price can go to any of the remaining 9 students.

The third price can go to any of the remaining 8 students.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 10 different objects in 3 places is

P(10,3) =

= = = 720.

Therefore, there are 10 × 9 × 8 = 720 ways to win first three prizes.

To find: number of ways of arranging 5 subjects in 6 periods.

Condition: at least 1 period for each subject.

5 subjects in 6 periods can be arranged in P (6,5).

Remaining 1 period can be arranged in P (5,1)

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Total arrangements = P(6,5) × P(5,1) = ×

= × = 720 × 5 = 3600.

Total number of ways is 3600 ways.

To find: number of ways of hanging 6 pictures on 4 picture nails.

There are 6 pictures to be placed in 4 places.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 6 different objects in 4 places is

P(6,4) = = = = 360

This can be done by 360 ways.

There are 8 alphabets in the word EQUATION.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 8 different objects in 8 places is

P(8,8) = = = = 40320

Hence there are 40320 words formed.

To find: 4 lettered word from letters of word NUMBERS

There are 7 alphabets in word NUMBERS.

The word is a 4 different letter word.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 7 different objects in 4 places is

P(7,4) = = = = 840.

Hence, they can be arranged in 840 words.

There are 6 letters in the word SUNDAY.

Different words formed using 6 letters of the word SUNDAY is P(6,6)

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 6 different objects in 6 places is

P(6,6) = = = = 720.

720 words can be formed using letters of the word SUNDAY.

When a word begins with D.

Its position is fixed, i.e. the first position.

Now rest 5 letters are to be arranged in 5 places.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 5 different objects in 5 places is

P(5,5) = = = = 120.

Therefore, the total number of words starting with D are 120.

To find: number of words starting with C and end with Y

There are 8 letters in word COURTESY.

Here the position of the letters C and Y are fixed which is 1^st and 8^th.

Rest 6 letters are to be arranged in 6 places which can be done in P(6,6).

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 6 different objects in 6 places is

P(6,6) = = = = 720.

Therefore, total number of words starting with C and ending with Y is 720.

There are 7 letters in the word ENGLISH.

Permutation of 7 letters in 7 places can be done in P(7,7) ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 7 different objects in 7 places is

P(7,7) = = = = 5040.

Hence, the total number of permutations is P 5040.

To find a number of words starting with E and ending with I, let us consider their position which is 1^st and 7^th.

The rest 5 letters are to be arranged in 5 places which can be done in P (5,5)

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 5 different objects in 5 places is

P(5,5) = = = = 120.

Therefore, there are 120 words starting with E and ending with I.

There are 7 letters in the word HEXAGON.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 7 different objects in 7 places is

P(7,7) = = = = 5040.

They can be permuted in P (7,7) = 5040 ways.

The vowels in the word are E, A, O.

Consider this as a single group.

Now considering vowels as a single group, there are total 5 groups (4 letters and 1 vowel group) can be permuted in P (5,5)

Now vowel can be arranged in 3! ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, the arrangement of 5 groups and vowel group is

P(5,5)×3! = = × 3! = × 6 = 720.

Hence total number of arrangements possible is 720.

To find: number of words formed

Condition: vowels occupy odd places

There are 8 letters in the word ORIENTAL and vowels are 4 which are O, I, E,A respectively.

There is 4 odd places in which 4 vowels are to be arranged.

The rest 4 letters can be arranged in 4! Ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, the total arrangement is

P(4,4) × 4! = × 4! = ×4! = × 24 = 576.

Therefore, total number of words formed are 576.

To find: number of words

Condition: consonants occupy odd places

There are total of 7 letters in the word FAILURE.

There are 3 consonants, i.e. F, L, R which are to be arranged in 4 places.

The rest 5 letters can be arranged in 4! Ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, the total nuber of words are

P(4,3) ×4! = ×4! = ×4! = ×24 = 576.

Hence total number of arrangements is 576.

To find: number of words

Condition: vowels should never occur together.

There are 6 letters in the word GOLDEN in which there are 2 vowels.

Total number of words in which vowels never come together =

Total number of words – total number of words in which the vowels come together.

A total number of words is 6! = 720 words.

Consider the vowels as a group.

Hence there are 5 groups that can be arranged in P(5,5) ways, and vowels can be arranged in P(2,2,) ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Total arrangements = P(5,5) × P(2,2) = ×

= × = 120 × 2 = 240.

Hence a total number of words having vowels together is 240.

Therefore, the number of words in which vowels don’t come together is 720 – 240 = 480 words.

To find: number of words

Condition: vowels occupy odd positions.

There are 7 letters in the word MACHINE out of which there are 3 vowels namely A C E.

There are 4 odd places in which 3 vowels are to be arranged which can be done P(4,3).

The rest letters can be arranged in 4! ways

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, the total number of words is

P(4,3)4!× = ×4!

= ×4! = ×24= 576.

Hence the total number of word in which vowel occupy odd positions only is 576.

(i) There is no restriction on letters

The word VOWELS contain 6 letters.

The permutation of letters of the word will be 6! = 720 words.

(ii) Each word begins with

Here the position of letter E is fixed.

Hence, the rest 5 letters can be arranged in 5! = 120 ways.

(iii) Each word begins with O and ends with L

The position of O and L are fixed.

Hence the rest 4 letters can be arranged in 4! = 24 ways.

(iv) All vowels come together

There are 2 vowels which are O ,E.

Consider this group.

Therefore, the permutation of 5 groups is 5! = 120

The group of vowels can also be arranged in 2! = 2 ways.

Hence the total number of words in which vowels come together are 120×2 = 240 words.

(v) All consonants come together

There are 4 consonants V,W,L,S. consider this a group.

Therefore, a permutation of 3 groups is 3! = 6 ways.

The group of consonants also can be arranged in 4! = 24 ways.

Hence, the total number of words in which consonants come together is 6×24 = 144 words.

For a number to be divisible by 5, the last digit should either be 5 or 0.

In this case, 5 is only possible.

For a four digit number to be between 3000 to 4000, in this case, should start with 3.

Therefore, the other 2 digits can be arranged by 4 numbers in P(4,2)

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of 4 different objects in 2 places is

P(4,2) =

= = = 12.

Therefore, there are 12 numbers present between 3000 to 4000 formed by using numbers 3,1,5,6,7,8.

Candidates in mathematics are not sitting together = total ways – the

Students are appearing for mathematic sit together.

The total number of arrangements of 8 students is 8! = 40320

When students giving mathematics exam sit together, then consider

them as a group.

Therefore, 6 groups can be arranged in P(6,6) ways.

The group of 3 can also be arranged in 3! Ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, total arrangments are

P(6,6) × 3! = ×3!

= ×3! = ×6 = 4320.

The total number of possibilities when all the students giving

mathematics exam sits together is 4320 ways.

Therefore, number of ways in which candidates appearing

mathematics exam is 40320 – 4320 = 36000.

(i) two of them, Rajan and Tanvy, are always together

Consider Rajan and Tanvy as a group which can be arranged in 2! = 2 ways.

The 3 children with this 1 group can be arranged in 4! = 24 ways.

The total number of possibilities in which they both come together is 2×24 = 48 ways.

(ii) two of them, Rajan and Tanvy, are never together

Two of them are never together = total number of possible ways of sitting – total number of ways in which they sit together.

A total number of possible way of arrangement of 5 students is 5! = 120 ways.

Therefore, the total number of arrangement when they both don’t sit together is = 120 – 48 = 72.

For the first row:

There are 7 teachers in which the position of principal is fixed.

Therefore, the teachers can be arranged in p(7,7) = 5040.

For the second row:

The tallest students are at the ends and can be arranged in 2! = 2 ways.

Rest 18 students can be arranged in P(18,18) ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, permutation of 18 different objects in 18 places is

P(18,18) =

= = = 18!

Therefore, a total number of arrangements of the second row is 2 × 18!

Total arrangements = 2 × 18! × 5040 = 10080 × 18!

The total number of arrangements is 10080×18!

In this question, n girls are to be seated alternatively between m boys.

There are m+1 spaces in which girls can be arranged.

The number of ways of arranging n girls is P(m+1,n) = ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, permutation of n different objects in m+1 places is

P(m+1,n) =

=

The arrangement of m boys can be done in P(m,m) ways.

Formula:

Number of permutations of n distinct objects among r different places, where repetition is not allowed, is

P(n,r) = n!/(n-r)!

Therefore, a permutation of m different objects in m places is

P(m,m) = = = m!

Therefore the total number of arrangements is .

To find: number of permutations of the letters of each word

Number of permutations of n distinct letters is n!

Number of permutations of n letters where r letters are of one kind, s letters of another kind, t letters of a third kind and so on =

(i) Here n = 5

P is repeated twice

So the number of permutations =

(ii) Here n = 7

A is repeated twice, and R is repeated twice

So, the number of permutations =

(iii) Here n = 8

M and E are repeated twice

So, the number of permutations =

(iv) Here n = 9

I is repeated twice, T is repeated thrice

So, the number of permutations =

(v) Here n = 11

E, N is repeated thrice, I,G are repeated twice

So the number of permutations =

(vi) Here n = 12

I and T are repeated twice, E is repeated thrice

So, the number of permutations =

To find: number of ways the letters can be arranged

The following table shows the possible arrangements

However, we see that case 1 = is the same as case 4 =

Similarly (case2,case 5), (case 3,case 6) are the same

So there are only 3 distinct cases

Hence the letters can be arranged in 3 distinct ways

To find: no of ways in which the balls can be arranged in a row where some balls are of the same kind

Total number of balls = 3+4+5 = 12

3 are of 1 kind, 4 are of another kind, 5 are of the third kind

Number of ways =

They can be arranged in 27720 ways

To find: number of 3 digit numbers he can make

If all were distinct, he could have made 3! = 6 numbers

But 2 number are the same

So the number of possibilities =

He can make 3 three - digit numbers using them

To find: Number of distinct signals possible

Total number of fags = 7

2 are of 1 kind, 3 are of another kind, and 2 are of the 3^rd kind

number of distinct signals =

Hence 210 different signals can be made

To find: number of words where vowels are together

Vowels in the above word are: A,A,E,E

Consonants in the above word: R,R,N,G,M,N,T

Let us denote the all the vowels by a single letter say Z

the word now has the letters, R,R,N,G,M,N,T,Z

R and N are repeated twice

Number of permutations =

Now Z is comprised of 4 letters which can be permuted amongst themselves

A and E are repeated twice

number of permutations of Z =

Total number of permutations =

The number of words that can be formed is 60480

To find: Number of words that can be formed so that vowels are never together

Number of words such that vowels are never the together = Total number of words - Number of words where vowels are together

Total number of words =

To find a number of words where vowels are together

Let the vowels I, I, A be represented by a single letter Z

the new word is NDZ

A number of permutations = 3! = 6

Z is composed of 3 letters which can be permuted amongst themselves.

Number of permutations of Z =

Number of words where vowels are together = = 18

Number of words where vowels are not together = 60 - 18 = 42

There are 42 words where vowels are not together

To find: number of arrangements without changing the relative position

The following table shows where the vowels and consonants can be placed

Consonants can be placed in the blank places

There are 3 spaces for vowels

There are 3 vowels out of which 2 are alike

Vowels can be placed in 3 ways

There are 4 consonants, and they can be placed in = 24 ways

Total number of arrangements = = 72 ways

72 arrangements can be made

To find: number of words which start and end with S

There are 4 places to fill up with 4 letters out of which 2 are of the same kind

Number of words =

12 words are possible

To find: number of words where L do not come together

Let the three L’s be treated as a single letter say Z

Number of words with L not the together = Total number of words - Words with L’s together

The new word is PARAEZ

Total number of words =

Words with L together =

Words with L, not together = 3360 - 720 = 2640

There are 2640 words where L do not come together

To find: number of words such that C and T are never together

Number of words where C and T are never the together = Total numbers of words - Number of words where C and T are together

Total number of words =

Let C and T be denoted by a single letter Z

New word is APAINZ

This can be permuted in = 360 ways

Z can be permuted among itself in 2 ways

number of words where C and T are together = 360 × 2 = 720

number of words where C and T are never together = 2520 - 720 = 1800

There are 1800 words where C and T are never together

To find: number of ways letters can be arranged such that all S’s are together

Let all S’s be represented by a single letter Z

New word is AAINATIONZ

Number of arrangements =

Letters can be arranged in 151200 ways

(i) There are 11 letters of which 2 are of 1 kind, 2 are of another kind, 2 are of the 3^rd kind

Total number of arrangements =

(ii)

There are 10 spaces to be filled by 10 letters of which 2 are of 3 different kinds

Number of arrangements = 453600

(iii)

There are 10 spaces to be filled by 10 letters of which 2 are of 2 different kinds

Number of arrangements =

(i)

There are 6 even places and 6 vowels out of which 2 are of 1 kind, 3 are of the 2^nd kind

The vowels can be arranged in 60

There are 6 consonants out of which 2 is of one kind

Number of permutations = 360

Total number of words =

(ii)

There are 6 vowels to arrange in

There are 6 consonants which can be arranged in

Total number of ways = 21600

(i) There are 11 letters of which 3 are of 1 kind, 2 are of the 2^nd kind ,3 are of the 3^rd kind

Number of arrangements = 554400

(ii) Let all the three T’s be denoted by a single letter Z

New word is INSIUIONZ

Number of permutations = 30240

(iii)

There are 9 places to be filled by 9 letters of which 3 are of 2 different kinds

Number of permutations =

To find: Number of 5 - digit numbers that can be formed

2 numbers are of 1 kind, and 2 are of another kind

Total number of permutations =

30 number can be formed

The table shows the places where the odd digits can be placed

There are 4 places

And 3 odd digits out of which 2 are of the same kind

Choose any 3 places out of the four places in ⁴C₃ ways = 4 ways

In each way, the 3 digits can be placed in ways = 3 ways

total number of ways in which odd digits occupy odd places = = 12

Now there are 4 remaining digits out of which 2 are same of 1 kind, and 2 are same as another kind

they can be arranged in the remaining places in = 6 ways

total number of numbers where odd digit occupies odd places = 12 × 6 = 72

There are 72 such numbers

To find: number of 7 digit

0 can not be in the first place because that would make a 6 digit number

Total number of 7 - digit numbers = Total number of number possible - Number of numbers with 0 at the first place

Total number of numbers possible =

Number of numbers with 0 at first place =

Number of 7 - digit numbers = 420 - 60 = 360

360 seven - digit numbers are possible

To find: number of 6 digit

0 cannot be in the first place because that would make a 5 - digit number

Total number of 6 - digit numbers = Total number of number possible - Number of numbers with 0 at the first place

Total number of numbers possible =

Number of numbers with 0 at first place =

Number of 6 - digit numbers = 180 - 30 = 150

150 six - digit numbers are possible

Alphabetical arrangement of letters: A,D,I,N

1^st word: ADIIN

To find other words:

Case 1: words starting with A

Number of words =

13^th word starts with D and is DAIIN

Case 2: words starting with D

Number of words =

Case 3: Words starting with I

Number of words =

(12+12+24+1)^th = 49^th word starts with N and is NAIID

Case 4: Words starting with N

Number of words =

(48+12)^th word is the last word which starts with N

60^th word = NDIIA

1st word: ADIIN

13^th word: DAIIN

49^th word: NAIID

60^th word: NDIIA

The first marble can be put into the pockets in 6 ways,

i.e. Choose 1 Pocket From 6 by ⁶C₁=6

Similarly second, third, Fourth, fifth & Sixth marble. Thus, the number of ways in which the child can put the marbles is 6⁵

As there is 5 banana, So suppose it as B₁, B₂, B₃, B₄, B₅ And Let the Boy be A₁, A₂, A₃

So B₁ can Be distributed to 3 Boys (A₁, A₂, A₃) by 3 ways,

Similarly, B₂, B₃, B₄, B₅ Can be distributed to 3 Boys by 3⁴

So total number of ways is 3⁵

Let Suppose Letterbox be B₁, B₂ and letters are L₁, L₂, L₃

So L₁ can be posted in any 2 letterboxes (B₁, B₂) by 2 ways

Similarly, L₂ can be posted in any 2 letterbox(B₁, B₂) by 2 ways

Similarly, L₃ can be posted in any 2 letterbox(B₁, B₂) by 2 ways

So total number of ways is 2³ = 8

Let Suppose 3 digit number as 3 boxes as shown below.First Box is at 100^thplace, the Second box is at 10^th place, and the Third box be at 1^st place.

To make a 3 digit number,

1^st box can be filled with nine numbers(1, 2, 3, 4, 5, 6, 7, 8, 9) if we include 0 in 1^st box then it become 2 digit number(i.e 010 is 2 digit number not 3 digit)

2^nd box can be filled with ten numbers(1, 2, 3, 4, 5, 6, 7, 8, 9, 0) as repetation is allowed.

Similarly 3^rd box can be filled with ten numbers(1, 2, 3, 4, 5, 6, 7, 8, 9, 0)

Total number of ways is 9 × 10 × 10 = 900

Let Suppose 4 digit number as 4 boxes as shown below.First Box is at 1000^thplace, the Second box is at 100^th place, the Third box is at 10^th place, and Fourth box is at 1^st place.

The 1^st box can be filled with four numbers(2, 3, 4, 5) if we include 0 in the 1^st box then it becomes 3 digit number(i.e. 0234 is 3 digit number, not 4 digits)

The 2^nd box can be filled with five numbers(0, 2, 3, 4, 5) as repetition is allowed.

Similarly, the 3^rd box can be filled with five numbers(0, 2, 3, 4, 5) as repetition is allowed.

Similarly, the 4^th box can be filled with five numbers(0, 2, 3, 4, 5) as repetition is allowed.

Total number of ways is 4 × 5 × 5 × 5 = 500

Let suppose 4 prizes be P₁, P₂, P₃, P₄ and 3 boys be B₁, B₂, B₃

Now P₁ can be distributed to 3 boys(B₁, B₂, B₃) by 3 ways,

Similarly, P₂ can be distributed to 3 boys(B₁, B₂, B₃) by 3 ways,

Similarly, P₃ can be distributed to 3 boys(B₁, B₂, B₃) by 3 ways,

And P₄ can be distributed to 3 boys(B₁, B₂, B₃) by 3 ways

So total number of ways is 3 × 3 × 3 × 3 =81

Let suppose 4 candidates be C₁, C₂, C₃, C₄ and 5 men be M₁, M₂, M₃, M₄, M₅

Now M₁ choose any one candidates from four (C₁, C₂, C₃, C₄) and give the vote to him by any 4 ways

Similarly, M₂ choose any one candidates from four (C₁, C₂, C₃, C₄) and give the vote to him by any 4 ways

Similarly, M₃ choose any one candidates from four (C₁, C₂, C₃, C₄) and give the vote to him by any 4 ways

Similarly, M₄ choose any one candidates from four (C₁, C₂, C₃, C₄) and give the vote to him by any 4 ways

And M₅ choose any one candidates from four (C₁, C₂, C₃, C₄) and give the vote to him by any 4 ways

So total numbers of ways are 4 × 4 × 4 × 4 × 4=1024

(i) Let choose 1 person from 6 by ⁶C₁=6 and arranged it in line

Now choose another person from remaining 5 by ⁵C₁=5 and arranged it in line

Similarly, choose another person from remaining 4 by ⁴C₁=4 and arranged it in line

Similarly, choose another person from remaining 3 by ³C₁=3 and arranged it in line

Similarly, choose another person from remaining 2 by ²C₁=2 and arranged it in line

And choose another person from remaining 1 by ¹C₁=1 and arranged it in line

So total number of ways is 6! =720

(ii) It is the same as above, by converting line arrangement into the circle but you need to remove some arrangement

Let suppose 6 persons as A, B, C, D ,E, F you need to arrange this 6 persons into a circle.

First, we arranged 6 persons in line(number of ways = 6!)

NOTE: A, B, C, D ,E, F and B, C, D, E, F, A consider as a different line, but when we arranged this 2 combination in circle then it becomes same,

i.e. Let takes us an example we need to arrange A, N, O, D, E.

We arrange it as shown. When we rotate first one, then 1^st and 2^nd became identical and so on that’s why all 5 are identical, and we count it as 1

Now come back to our questions

So total number of arrangement is (6-1)! = 5! = 120

NOTE: When you want to arrange n persons in circle then a total number of ways is n!/n,

i.e. Total number of ways = (n-1)!

Let first arranged 5 men in the round table by 4! (by using the formula (n-1)! Mention above)

Now there are 5 gaps created between 5 men (check the figure)

So we arrange 5 ladies in this gap by 5!

A total number of ways to arrange 5 men and 5 ladies is 5! × 4! = 2880

First assume the president(P), Joint secretary(JS) and secretary(S) to be 1 members(as shown below)

So there are 9 members, a number of ways to arrange this 9 people is 8! (the formula used (n-1)!)

Now we need to look at the internal arrangement. There are 2 arrangement possible

So total number of arrangement are (8!) × 2 = 80,640

By using the formula (n-1)! (mention in Solution-1)

So 8 persons can be arranged by 7!

Now each person have the same neighbours in the clockwise and anticlockwise arrangement

Total number of arrangement are (7!)/2 = 2520

We know that necklace in the form of a circle, So we need to arrange 20 pearls in Circle

20 pearls can be arranged by 19!

Now each pearl have the same neighbours in the clockwise and anticlockwise arrangement

Total number of arrangement are (19!)/2

It is also in the form of a circle, So we need to arrange 16flowers in Circle

16 flowers can be arranged by 15!

Now each flower have the same neighbours in the clockwise and anticlockwise arrangement

Total number of arrangement are (15!)/2

To Find: Value of n

Given: (n+1)! = 12× [(n-1)!]

Formula Used: n! = (n) × (n-1) × (n-2) × (n-3) ………. 3 × 2 × 1

Now, (n+1)! = 12× [(n-1)!]

⇒ (n+1) × (n) × [(n-1)!] = 12 × [(n-1)!]

⇒ (n+1) × (n) = 12

⇒ n²+n = 12

⇒ n²+n-12 = 0

⇒ (n-3) (n+4) = 0

⇒ n = 3 or, n = -4

But, n=-4 is not possible because in case of factorial (!) n can not be negative.

Hence, n=3 is the correct answer.

To Find: Value of n

Given:

Formula Used: n! = (n) × (n-1) × (n-2) × (n-3) ………. 3 × 2 × 1

Now,

(4! = 24, 5! = 120)

⇒ x = 36

Given: We have 10 numbers i.e. 0,1,2,3,4,5,6,7,8,9

To Find: Number of 3-digit numbers formed with no repetition of digits.

Conditions: No digit is repeated

Let us represent the 3-digit number

First place can be filled with 9 numbers, i.e. 1,2,3,4,5,6,7,8,9 (0 cannot be placed as it will make it a 2-digit number) = 9 ways

Second place can be filled with remaining 9 numbers (as one number is used already) = 9 ways

Similarly, third place can be filled with 8 numbers = 8 ways

Total number of 3-digit numbers which can be formed

= 9 × 9 × 8= 648

Given: We have 5 digits i.e. 2,3,4,5,6

To Find: Number of 3-digit numbers

Condition: (i) Number should be greater than 600

(ii) Repetition of digits is allowed

For forming a 3 digit number, we have to fill 3 vacant spaces.

But as the number should be above 600, hence the first place must be occupied with 6 only because no other number is greater than 6.

Let us represent the 3-digit number

So the first place is filled with 6 = 1 ways

Second place can be filled with 5 numbers = 5 ways

Third place can be filled with 5 numbers = 5 ways

Total number of ways = 1 × 5 × 5 = 25

Total number of 3-digit numbers above 600 which can be formed by using the digits 2, 3, 4, 5, 6 with repetition allowed is 25

Given: We have 5 digits, i.e. 4,5,6,7,8

To Find: Number of numbers divisible by 5

Condition: (i) Number should be between 4000 and 5000

(ii) Repetition of digits is allowed

Here as the number is lying between 4000 and 5000, we can conclude that the number is of 4-digits and the number must be starting with 4.

Now, for a number to be divisible by 5 must ends with 5

Let us represent the 4-digit number

Therefore,

The first place is occupied by 4 = 1 way

The fourth (last) place is occupied by 5 = 1 way

The second place can be filled by 5 numbers = 5 ways

The third place can be filled by 5 numbers = 5 ways

Total numbers formed = 1 × 5 × 5 × 1 = 25

There are 25 numbers which are divisible by 5 and lying between 4000 and 5000 and can be formed from the digits 4, 5, 6, 7, 8 with repetition of digits.

Given: We have 6 letters

To Find: Number of words formed with Letter of the word ‘CHEESE.’

The formula used: The number of permutations of n objects, where p₁ objects are of one kind, p₂ are of the second kind, ..., p_k is of a k^th kind and the rest if any, are of a different kind is

Suppose we have these words – C,H,E₁,E₂,S,E₃

Now if someone makes two words as CHE₁E₃SE₂ and CHE₂E₃SE₁

These two words are different because E₁, E2 and E₃ are different but we have three similar E’s hence, in our case these arrangements will be a repetition of same words.

In the word CHEESE, 3 E’s are similar

∴ n = 6, p₁ = 3

In 120 ways the letters of the word ‘CHEESE’ can be arranged.

Given: We have 12 letters

To Find: Number of words formed with Letter of the word ‘PERMUTATIONS.’

The formula used: The number of permutations of n objects, where p₁ objects are of one kind, p₂ are of the second kind, ..., p_k is of a k^th kind and the rest if any, are of a different kind is

In the word ‘PERMUTATIONS’ we have 2 T’s.

We have to start the word with P and end it with S, hence the first and last position is occupied with P and S respectively.

As two positions are occupied the remaining 10 positions are to be filled with 10 letters in which we have 2 T’s.

NOTE:- Unless specified , assume that repetition is not allowed.

Let us represent the arrangement

Hence,

The first place is occupied by P = 1 way

The last place (12^th) is occupied by S = 1 way

For the remaining 10 places:

Using the above formula

Where,

n=10

p₁=2

Total number of ways are ways.

In 1814400 ways the letters of the word ‘PERMUTATIONS’ can be arranged if each word starts with P and ends with S.

Given: We have 9 letters

To Find: Number of words formed with Letter of the word ‘ALLAHABAD.’

The formula used: The number of permutations of n objects, where p₁ objects are of one kind, p₂ are of the second kind, ..., p_k is of a k^th kind and the rest if any, are of a different kind is

‘ALLAHABAD’ consist of 9 letters out of which we have 4 A’s and 2 L’s.

Using the above formula

Where,

n=9

p₁=4

p₂=2

7560 different words can be formed by using all the letters of the word ‘ALLAHABAD.’

Given: We have 5 letters

To Find: Number of words formed with Letter of the word ‘APPLE.’

The formula used: The number of permutations of n objects, where p₁ objects are of one kind, p₂ are of the second kind, ..., p_k is of a k^th kind and the rest if any, are of a different kind is

‘APPLE’ consists of 5 letters out of which we have 2 Ps.

Using the above formula

Where,

n=5

p₁=2

There are 60 permutations of the letters of the word ‘APPLE’.

Given: We have 6 letters

To Find: Number of words formed with Letter of the word ‘SUNDAY.’

‘SUNDAY’ consist of 6 letters.

NOTE: - Unless specified, assume that repetition is not allowed.

Let us represent the arrangement with an example

6 ways5 ways4 ways3 ways2 ways1 way

We have 6 places

First place can be filled with 6 letters, i.e. S,U,N,D,A,Y = 6 ways

Second place can be filled with 5 letters (as one letter is already used in the first place) = 5 ways

Similarly,

Third place can be filled with 4 letters = 4 ways

The fourth place can be filled with 3 letters = 3 ways

The fifth place can be filled with 2 letters = 2 ways

The sixth place can be filled with 1 letters = 1 ways

Total number of letters = 6 × 5 × 4 × 3 × 2 × 1 = 720

720 words can be formed by the letters of the word ‘SUNDAY.’

Given: We have 4 letters and 5 letter boxes

To Find: Number of ways of posting letters.

One letter can be posted in any of 5 letter boxes.

We have to assume that all the letters are different.

So for first letter i.e. , we have 5 ways

Similarly for,

Total number of ways = 5 × 5 × 5 × 5 = 625

In 625 ways 4 letters can be posted in 5 letter boxes.

Given: We have 4 women and 4 taps

To Find: Number of ways of drawing water

Condition: No tap remains unused

Let us represent the arrangement

The first woman can use any of the four taps

The second woman can use the remaining three taps = 3 ways

The third woman can use the remaining two taps = 2 ways

The fourth woman can use the remaining one tap = 1 way

Total number of ways = 4 × 3 × 2 × 1 = 24

There is 24 number of ways in which 4 women can draw water from 4 taps such that no tap remains unused.

Given: We have 3 digits, i.e. 0, 1 and 2

To Find: Number of 5-digit numbers formed

Let us represent the arrangement

For forming a 5-digit number, we have to fill 5 vacant spaces.

But the first place cannot be filled with 0, hence for filling first place, we have only 1 and 2

First place can be filled with 2 numbers, i.e. 1, 2 = 2 ways

Second place can be filled with 3 numbers = 3 ways

Third place can be filled with 3 numbers = 3 ways

The fourth place can be filled with 3 numbers = 3 ways

The fifth place can be filled with 3 numbers = 3 ways

Total number of ways = 2 × 3 × 3 × 3 × 3 = 162

162 5-digit numbers can be formed by using the digits 0, 1 and 2.

Given: We have 5 boys and 3 girls

To Find: Number of ways of seating so that 5 boys and 3 girls are seated in a row and each girl is between 2 boys

The formula used: The number of permutations of n different objects taken r at a time (object does not repeat) is

The only arrangement possible is

B__B__B__B__B

Number of ways for boys

=120

There are 3 girls, and they have 4 vacant positions

Number of ways for girls

=24

Total number of ways = 24 × 120 = 2880

In 2880 ways 5 boys and 3 girls can be seated in a row so that each girl is between 2 boys.

Given: We have toys with bearing 4, 4 and 5

To Find: Number of 3-digit numbers he can make

The formula used: The number of permutations of n objects, where p₁ objects are of one kind, p₂ are of the second kind, ..., p_k is of a k^th kind and the rest, if any, are of a different kind is

The child has to form a 3-digit number.

Here the child has two 4’s.

We have to use the above formula

Where,

n=3

p₁=2

The numbers are 544, 454 and 445.

He can make 3 3-digit numbers.

Given: We have 6 letters

To Find: Number of ways to arrange letters P,E,N,C,I,L

Condition: N is always next to E

Here we need EN together in all arrangements.

So, we will consider EN as a single letter.

Now, we have 5 letters, i.e. P,C,I,L and ‘EN’.

5 letters can be arranged in

⇒

⇒

⇒ 120

In 120 ways we can arrange the letters of the word ‘PENCIL’ so that N is always next to E