Eight chairs are numbered 1 to 8. Two women and 3 men wish to occupy one chair each. First the women choose the chairs from amongst the chairs 1 to 4 and then men select from the remaining chairs. Find the total number of possible arrangements.
Formula:-
nPr
Given: -
W1 can take chairs marked 1 to 4 in 4 different way.
W2 can 3 chairs from marked 1 to 4 in 3 different ways.
So, total no of ways in which women can take seat
4P2
4P2
=12
There will be 6 chairs remains
M1 take seat in any of the 6 chairs in 6 different ways,
M2 can take seat in any of the remaining 5 chairs in 5 different ways
M3 can take seat in any of the remaining 4 chairs in 4 different ways.
So, total no of ways in which men can take seat
6P3
=120
Hence total number of ways in which men and women can be seated
4P2x6P3
=1440
If the letters of the word RACHIT are arranged in all possible ways as listed in dictionary. Then what is the rank of the word RACHIT?
[Hint: In each case number of words beginning with A, C, H, I is 5!]
Given:-
we arrange in alphabetical order
we get A C H I
number of Words start with A=5!
number of Words start with C=5!
number of Words start with H=5!
number of Words start with I=5!
number of Words start with R is RACHIT
= 5! +5!+5!+5!
120+120+120+120=480
Hence number of words with R is RACHIT
Thus 480+1=481
A candidate is required to answer 7 questions out of 12 questions, which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. Find the number of different ways of doing questions.
Formula: -(i) nCr
No of questions in groupA=6
No of questions in group B=6
The different ways of doing the questions are
(6C2x6C5)+(6C3x6C4)+(6C4x6C3)+(6C5x6C2)
=(15×6)+(20×15)+(15×20)+(6×15)
=780
Out of 18 points in a plane, no three are in the same line except five points which are collinear. Find the number of lines that can be formed joining the point.
[Hint: Number of straight lines =18C2 – 5C2 + 1]
Formula:- (i)nCr
Given:-
Number of points =18
Number of Collinear points =5
Number of lines form by 18 points =18C2
For 5 points are collinear=5C2
the number of lines that can be formed joining the point.
=18C2-5C2+1
=153-10+1
=144
We wish to select 6 persons from 8, but if the person A is chosen, then B must be chosen. In how many ways can selections be made?
Formula:- (i)nCr
Given: -
Case 1: If A and B both are selected
=1x1x6C4
Case 2:If neither A nor B are selected
6C6
If B is selected but A is not selected
1x6C5
Adding above
15+1+6=22
How many committee of five persons with a chairperson can be selected from 12 persons. [Hint: Chairman can be selected in 12 ways and remaining in 11 C4.]
Formula:- (i)nCr
A chairperson can be selected in =12
The 4 other person can be selected by
=11C4
Selection of 5 people can be done in
=330 x 12
=3960
How many automobile license plates can be made if each plate contains two different letters followed by three different digits?
Given:-
Number of letters in automobile license plates =2
We know that there are 26 alphabets
so, without repetition letter can be arranged
= 26×25
=650
Number of digitsin automobile license plates =3
We know that there 10 digits
Hence the no of digits with no repetitions
=10×9×8=720
Total number of way automobile license plates
=720×650
=468000
A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected from the lot.
Formula: -(i)nCr
Given: -Number of black ball =5,Number of red ball =6
Number of ways of selection of 2 black balls
=5C2
Number of ways of selection of 3 red balls
=5C3
Number of ways of selection of 2 black & 3 red ball
=5C2x5C3
=10×20
=200
Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.
permutations of n distinct things taken r together
=nCr
And when 3 particular things must occur together= n-3Cr-3
= n-3Cr-3× (r – 2)! × 3!
Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together.
Formula:- (i)nPr
Given:-Total number of vowels letter =3, Total no of consonants letter =5
The vowels can be placed in
6P3
The number of way consonants can be arranged placed
=5! =120
Total number of ways it can be arranged
=5!×6P3=120×120
=14400
Find the number of positive integers greater than 6000 and less than 7000 which are divisible by 5, provided that no digit is to be repeated.
(i)Thousand Place is to be fill with 6. Number of way =1
Unit place can be fill either 0 or 5. Number of way=2
Hundred place can be fill with remaining 8 digits. Number of way =8
ten place can be fill with 7 digits. Number of ways= 7.
Thus, required number will be
=1 x 8 x 7 x 2
=112
There are 10 persons named P1,P2,P3, ... P10. Out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P1 must occur whereas P4 and P5 do not occur. Find the number of such possible arrangements. [Hint: Required number of arrangement =7C4× 5!]
Formula:- (i)nCr
Given:- there are 10 person named P1, P2, P3, ... P10.
Number of ways of P1 arrangement =5! =120
Number of ways arrangement of other
7C4
Therefore, required number of arrangement =
=35 x 120
=4200
There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.[Hint: Required number = 210 – 1].
Formula:- (i) nCr
(ii)
Given:- number of lamps in a hall =10
one of them can be switched on independently
Find the number of ways in which the hall can be illuminated
=
=210-1
=1024-1
=1023
A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box , if atleast one black ball is to be included in the draw.
[Hint: Required number of ways =3C1×6C2+3C2x 6C1+3C3.]
FORMULA:- (i)nCr
Number of ways drawing at least one black ball =
=( 1 black and 2 other )or( 2 black and 1 other )or (3 black)
3C1x6C2+3C2x6C1+3C3
=3×15+3×6+1
= 45 + 18 + 1
= 64
If nCr – 1 = 36, nCr = 84 and nCr + 1 = 126, then find rC2.
[Hint: From equation using to find the value of r.]
Formula:- (i)nCr
Given:-
nCr-1=36,nCr=84,nCr+1=126
2n-2r=3r+3
⇒2n-3=5r-------------------(i)
=3n-3r+3=7r
3n+3=10r-----------------(2)
From (1) and (2)
2(2n-3)=3n+3
4n-3n-6-3=0
n=9
and r=3
now
rC2=3C2
=3
Find the number of integers greater than 7000 that can be formed with the digits 3, 5, 7, 8 and 9 where no digits are repeated. [Hint: Besides 4-digit integers greater than 7000, five digit integers are always greater than 7000.]
GIVEN:-Digits to be use are 3,5,7,8,9
There should be 4 digits integers greater than 7000
And five digits integers are always greater than 7000
The number of integers is 5P5
For a four-digit integer to be greater than 7000 it must begin with 7,8 or 9.
The number of such integer is 3x4P3
=3x4P3
=3(24) =72
Total no of ways
=120+72
=192
If 20 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, in how many points will they intersect each other?
Let first line intersect point =0
2nd line intersect point=1
3rd line intersect point =2+1
4th line intersect point =3+2+1
…………………………………………….
………………………………………….
nth line intersect point =(n-1) +(n-2).....(3)(2)(1) where n=20
=1+2+3....(n−1)
=19×10
=190
In a certain city, all telephone numbers have six digits, the first two digits always being 41 or 42 or 46 or 62 or 64. How many telephone numbers have all si x digits distinct?
Given:- all telephone numbers have si x digits
the first two digits always being 41 or 42 or 46 or 62 or 64.
=5
First two digits should can be fill in 5 ways
The four-digit remaining will be filled in
=8P4
Therefore, telephone numbers have all six digits distinct
=5×1680
=8400
In an e x amination, a student has to answer 4 questions out of 5 questions; questions 1 and 2 are however compulsory. Determine the number of ways in which the student can make the choice.
Formula:- (i)nCr
Given:-number of questions =5
Number of questions have toanswered =4
Compulsory questions are 1 and 2
The number of ways in which the student can make the choice
=3C2
=3 ways
A convex polygon has 44 diagonals. Find the number of its sides. [Hint: Polygon of n sides has (nC2 – n) number of diagonals.]
Formula:-
(i)nCr
Let the polygon has n sides.
By joining any two vertices, we will get a side or a diagonal.
No. of line segments obtained by joining n vertices
=nC2
Out of these n line segments are sides of the polygon.
So, no. of diagonals of the polygon
=nC2-n=44
(n - 11) (n + 8) = 0
n = 11 or n = - 8
The polygon has 11sides.
18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?
Given:-number of mice =18,Number of groups =3
The groups are equally large hence each group can have =6 mice
The number of ways of placement of mice=18!
For each group placement of mice =6!
The required number way
A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if
(a) they can be of any colour
(b) two must be white and two red and
(c) they must all be of the same colour.
Formula:- (i)nCr
Given:-number of white marbles =6, number of red marbles =5
number of marbles = 6 white + 5 red = 11 marbles
(A)If they can be of any colour
Then we have to select 4 marbles out of 11
∴ Required number of ways
=11C4
(b)Number of ways of choosing two white and two white and two red are
=6C2x5C2
=15x10
=150
(c) If they all must be of same colour,
Four white marbles out of 6 can be selected
=6C4
And 4 red marbles out of 5 can be selected
=5C4
∴ Required number of ways
6C4+5C4=15+5=20
In how many ways can a football team of 11 players be selected from 16 players? How many of them will
(i) include 2 particular players?
(ii) exclude 2 particular players?
Formula:- (i)nCr
Given:- total number of player=11
11 players can be selected out of 16
=16C11
= 4368 ways.
(i)include 2 particular players
14C9
= 2002
(i) exclude 2 particular players
14C11
= 364
A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and atleast 5 from Class X II. If there are 20 students in each of these classes, in how many ways can the team be constituted?
Formula:- (i)nCr
A team of 11 students can be constituted in the following two ways:-
case(i):- 5 students from class X I and 6 from X II
case(ii):- 6 students from class X I and 5 from X II
the number ways can the team be constituted: -
Case(i)+case(ii)
=20C5.20C6+20C6.20C5
=2(20C5.20C6)
A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has
(i) no girls
(ii) at least one boy and one girl
(iii) at least three girls.
Formula:- (i)nCr
(i) No girls
Total number of ways
=4C0.7C5
=21
(ii) at least one boy and one girl
CASE(A)1 boy and 4 girls
=7C1.4C4
=7
CASE(B)2 boys and3 girls
=7C2.4C3
=84
CASE(C) 3boys and 2girls
=7C3.4C2
=210
CASE (D)4 boys and 1 girls
=7C4.4C1
=140
Total number of ways= CASE(A)+CASE(B)+CASE(C)+CASE(D)
=7+84+210+140
=441
(iii) At least three girls
=4C3.7C2+4C4.7C1
= 4 × 21 + 7
= 84 + 7
= 91
If nC12 = nC8, then n is equal to
A. 20
B. 12
C. 6
D. 30
Given:-nC12 = nC8
Formula: - (i) nCr
n-8=12
n=20
The number of possible outcomes when a coin is tossed 6 times is
A. 36
B. 64
C. 12
D. 32
possible outcomes when a coin is tossed 6 times is
The number of different four digit numbers that can be formed with the digits 2, 3, 4, 7 and using each digit only once is
A. 120
B. 96
C. 24
D. 100
Formula:- (i)nPr=
four digit numbers that can be formed with the digits 2, 3, 4, 7
and using each digit only once=4P4
The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time is
A. 432
B. 108
C. 36
D. 18
The sum of the digits in unit place of all the numbers formed with the help of 3, 4, 5 and 6 taken all at a time is
=(3+4+5+6)3!
=108
Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to
A. 60
B. 120
C. 7200
D. 720
Formula:- (i) nCr
Number of vowel=4
Number of consonant=5
Total number of words formed by 2 vowels and 3 consonants
=4C2.5C3
=
2 vowel and 3 consonant =5!
Total number of word
A five digit number divisible by 3 is to be formed using the numbers 0, 1, 2, 3, 4 and 5 without repetitions. The total number of ways this can be done is
A. 216
B. 600
C. 240
D. 3125
[Hint:5 digit numbers can be formed using digits 0, 1, 2, 4, 5 or by using digits 1, 2, 3, 4, 5 since sum of digits in these cases is divisible by 3.]
5-digit numbers can be formed using digits 0, 1, 2, 4, 5
4x4x3x2x1=96
5-digit numbers can be formed using digits 1, 2,3, 4, 5=5!
Total number of ways=5! +96
=216
Everybody in a room shakes hands with everybody else. The total number of hand shakes is 66. The total number of persons in the room is
A. 11
B. 12
C. 13
D. 14
Formula:- (i) nCr
Let total time of handshakes=nC2
⇒n2-n=132
⇒(n-12) (n+11)=0
n=12 or n=-11
The number of triangles that are formed by choosing the vertices from a set of 12 points, seven of which lie on the same line is
A. 105
B. 15
C. 175
D. 185
Formula:- (i) nCr
Triangle are form by 3 point
Total number of triangle formed
Triangle formed by vertices from a set of 12 points and seven point
12C3-7C3
=220-35
=185
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines is
A. 6
B. 18
C. 12
D. 9
Formula:- (i) nCr
The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines
=4C2.3C2
=6x3=18
The number of ways in which a team of eleven players can be selected from 22 players always including 2 of them and excluding 4 of them is
A.16C11
B.16C5
C.16C9
D.20C9
Formula:- (i)nCr=
GIVEN:- 22 players
players always including 2 of them and excluding 4 is
=22-2-4=16
Number of way of selection
=16C9
The number of 5-digit telephone numbers having atleast one of their digits repeated is
A. 90,000
B. 10,000
C. 30,240
D. 69,760
Formula:- (i)nPr=
Total of of 5-digit telephone numbers=
If no number is repeated=10P5
The number of 5-digit telephone numbers having atleast one of their digits repeated is
=105-10P5
The number of ways in which we can choose a committee from four men and six women so that the committee includes at least two men and exactly twice as many women as men is
A. 94
B. 126
C. 128
D. None
Formula:- (i)nCr=
Given:-
Number of women=6,number of men =4
Group I should of 2 men and 4 women=4C2.6C4
GROUP II should of 3 men and 6 women=4C3.6C6
Required committee=4C2.6C4+4C3.6C6
=90+4
=94
The total number of 9 digit numbers which have all different digits is
A. 10!
B. 9!
C. 9 × 9!
D. 10×10!
The place can filled by 9 way
Last places can be fill by 9! way
The total number of 9-digit numbers which have all different digits is
=9.9!
The number of words which can be formed out of the letters of the word ARTICLE, so that vowels occupy the even place is
A. 1440
B. 144
C. 7!
D.4C4 × 3C3
Formula:- (i)nPr=
Vowel at even places can be formed by=3P3
Consonants at odd places =4P4
Required arrangement =3! x4!
=144
Given 5 different green dyes, four different blue dyes and three different red dyes, the number of combinations of dyes which can be chosen taking at least one green and one blue dye is
A. 3600
B. 3720
C. 3800
D. 3600
[Hint: Possible numbers of choosing or not choosing 5 green dyes, 4 blue dyes and 3 red dyes are 25, 24 and 23, respectively.]
Chose 5 different green dye
Chose 4 blue dye
Chose 3 different green dye
Required selection is
Fill in the Blanks
If nPr = 840, nCr = 35, then r = ______.
Formula:-
(i)nPr
(ii) nCr
Given:-
nPr = 840,
nCr = 35
nPr
nCr
Diving both nPrand nCr
Fill in the Blanks
15C8 + 15C9 – 15C6 – 15C7 = ______.
Formula:-
(i)nCr=nCn-r
15C8+15C9-15C6-15C7
=15C15-8+15C15-9-15C6-15C7
=15C7+15C6-15C6-15C7=0
Fill in the Blanks
The number of permutations of n different objects, taken r at a line, when repetitions are allowed, is ______.
Fill in the Blanks
The number of different words that can be formed from the letters of the word INTERMEDIATE such that two vowels never come together is ______.
[Hint: Number of ways of arranging 6 consonants of which two are alike isand number of ways of arranging vowels
Number of vowel =6
Number of consonants =6
Arranging vowel =
Arranging consonant=
Total number of words
Fill in the Blanks
Three balls are drawn from a bag containing 5 red, 4 white and 3 black balls. The number of ways in which this can be done if at least 2 are red is ______.
Formula:-
(i) nCr
Given:-
Number of red=5
Number of white=4
Number of black=3
Required number of way=5C2.7C1+5C3
=180.
Fill in the Blanks
The number of six-digit numbers, all digits of which are odd is ______.
Number of digit=10
Number of odd digit=5
number of six-digit numbers=
Fill in the Blanks
In a football championship, 153 matches were played . Every two teams played one match with each other. The number of teams, participating in the championship is ______.
Formula:-
(i) nCr
Given:-
Number of team =n
Number of matches=nC2
nC2=153
⇒n2-n=153
⇒n2-n-153=0
⇒(n-18)(n+17)=0
⇒n=18,-17
n=18
Fill in the Blanks
The total number of ways in which six ‘+’ and four ‘–’ signs can be arranged in a line such that no two signs ‘–’ occur together is ______.
Formula:-
(i) nCr
Number of (-)=4
Number of (+)=6
Number of ways=7C4=35
Fill in the Blanks
A committee of 6 is to be chosen from 10 men and 7 women so as to contain atleast 3 men and 2 women. In how many different ways can this be done if two particular women refuse to serve on the same committee.
[Hint: At least 3 men and 2 women: The number of ways = 10C3 × 7C3 + 10C4 × 7C2.
For 2 particular women to be always there: the number of ways = 10C4 + 10C3 × 5C1.
The total number of committees when two particular women are never together = Total – together.]
Formula:-
(i) nCr
Number of men =10
Number of women=7
At least 3 men and 2 women: The number of ways = 10C3 × 7C3 + 10C4 × 7C2.
For 2 particular women to be always there: the number of ways = 10C4 + 10C3 × 5C1.
The total number of committees when two particular women are never together
= Total – together
=(10C3 × 7C3 + 10C4 × 7C2)-( 10C4 + 10C3 × 5C1)
=8610-810
=7800
Fill in the Blanks
A box contains 2 white balls, 3 black balls and 4 red balls. The number of ways three balls be drawn from the box if at least one black ball is to be included in the draw is ______.
Formula:-
(i) nCr
Number of white ball =2
Number of black ball=3
Number of red ball=4
selection of 3 balls if at least 1 black ball
=3C1.6C2+3C2.6C1+3C3
=45+18+1=64
State whether the statements in True or False.
There are 12 points in a plane of which 5 points are collinear, then the number of lines obtained by joining these points in pairs is 12C2 – 5C2.
FALSE
Required number of lines are 12C2 – 5C2 + 1
the number of lines obtained by joining these points in pairs is 12C2 – 5C2+1.
State whether the statements in True or False.
Three letters can be posted in five letterboxes in 35 ways.
FALSE
Each of the letter can be posted in any of the five letter boxes.
Three letters can be posted in five letterboxes in53=125
State whether the statements in True or False.
In the permutations of n things, r taken together, the number of permutations in which m particular things occur together is n–mPr–m × rPm.
FALSE
According to theorem:-In the permutations of n things, r taken together, the number of permutations in which m particular things occur together n-mPr-m
State whether the statements in True or False.
In a steamer there are stalls for 12 animals, and there are horses, cows and calves (not less than 12 each) ready to be shipped. They can be loaded in 312 ways.
TRUE
They can be loaded in 312
State whether the statements in True or False.
If some or all of n objects are taken at a time, the number of combinations is 2n–1.
TRUE
State whether the statements in True or False.
There will be only 24 selections containing at least one red ball out of a bag containing 4 red and 5 black balls. It is being given that the balls of the same colour are identical.
TRUE
24 selections containing at least
one red ball out of a bag containing 4 red and 5 black balls.
=5x5
=25
State whether the statements in True or False.
Eighteen guests are to be seated, half on each side of a long table. Four particular guests desire to sit on one particular side and three others on other side of the table. The number of ways in which the seating arrangementscan be made is .
[Hint: After sending 4 on one side and 3 on the other side, we have to select out of 11; 5 on one side and 6 on the other. Now there are 9 on each side of the long table and each can be arranged in 9! ways.]
true
After sending 4 on one side and 3 on the other side, we have to select out of 11
=11C5
5 on one side and 6 on the other.
=11-5C6=6C6
Now there are 9 on each side of the long table and each can be arranged in 9! ways.
The total arrangement =11C5x9!x6C6x9!
State whether the statements in True or False.
A candidate is required to answer 7 questions out of 12 questions which are divided into two groups, each containing 6 questions. He is not permitted to attempt more than 5 questions from either group. He can choose the seven questions in 650 ways.
false
6C2x6C5+6C3x6C4+6C4x6C3+6C5x6C2=780
State whether the statements in True or False.
To fill 12 vacancies there are 25 candidates of which 5 are from scheduled castes. If 3 of the vacancies are reserved for scheduled caste candidates while the rest are open to all, the number of ways in which the selection can be made is 5C3 × 20C9.
false
Select 3 sc candidate out of 5=5C3
Select 9 candidate out of 22=22C9
the number of ways in which the selection can be made is5C3 × 20C9.
There are 3 books on Mathematics, 4 on Physics and 5 on English. How many different collections can be made such that each collection consists of:
Given:-
Number of book of mathematics=3
Number of book of physics=4
Number of book of English=5
3C1.4C1.5C1=60
=(23-1)( 23-1)( 25-1)
=3225
Adding book other than English=4+3=7
=(25-1).27
3968
Answer is above in table
Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition:
Given:-
Number of boy=5
Number of girls=5
=(5!.5!)+(5!.5!)
=(5!)2+(5!)2
=5!.6!
=2!.5!.5!
Total number of boys and girls=5+5=10
Number of way=10!-5!.6!
Answer is above in table
There are 10 professors and 20 lecturers out of whom a committee of 2 professors and 3 lecturer is to be formed. Find :
Given:-
Number of professors=10
Number of lecturers=20
=10C2.20C3
=9C1.20C3
=10C2.19C2
=10C2.19C3
Answer is above in table
Using the digits 1, 2, 3, 4, 5, 6, 7, a number of 4 different digits is formed. Find
Given:-total number of digit to be form=4
=7P4=840
Number which are ending with 2
6P3=120
Number which are ending with 4
6P3=120
Number which are ending with 6
6P3=120
how many numbers are exactly divisible by 2=120+120+120=360
=40
The number which last two digit divisible by 4
Are 12,16,24,32,36,44,52,56,64,72,76
11P2.5P2
=200
Answer is above in table
How many words (with or without dictionary meaning) can be made from the letters of the word MONDAY, assuming that no letter is repeated, if
Given:-
Total number of letter of MONDAY=6
Total vowel=2
=6P4=360
=6P6=720
=2.5!
Answer is above in table