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NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

Edited By Ravindra Pindel | Updated on Sep 12, 2022 05:54 PM IST

NCERT Exemplar Class 11 Maths Chapter 7 solutions covers Permutations and Combinations. Permutations are explained as arrangements of objects, all taken at the same time or at different times, but in a definite order. Combination refers to the pairing of two or more objects with each other to know its suitability and results. NCERT Exemplar Class 11 Maths Solutions Chapter 7 cover arrangement and pairing over objects for the derivation of answers. Along with these two concepts, they have certain denotations that help in making up of formulas and understanding the way of formula derivation.

JEE Main Scholarship Test Kit (Class 11): Narayana | Physics Wallah | Aakash | Unacademy

Suggested: JEE Main: high scoring chapters | Past 10 year's papers

Scholarship Test: Vidyamandir Intellect Quest (VIQ)

This Story also Contains
  1. NCERT Exemplar Class 11 Maths Solutions Chapter 7: Exercise - 1.3
  2. Important Notes From NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations
  3. Main Subtopics in NCERT Exemplar Class 11 Maths Chapter 7 Solutions
  4. What will the students learn from NCERT Exemplar Class 11 Maths Chapter 7 Solutions?
  5. NCERT Solutions for Class 11 Mathematics Chapters
  6. Important Topics in NCERT Exemplar Class 11 Maths Chapter 7 Solutions Permutations and Combinations

NCERT Exemplar Class 11 Maths Solutions Chapter 7: Exercise - 1.3

Question:1

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.

Answer:

Let the women choose the chairs from amongst the chairs numbered 1 to 4.

Two women can be arranged in 4 chairs in ^4P_2 ways

In remaining 6 chairs, 3 men can be arranged in ^6P_3ways

Total possible arrangements= 4_{\mathrm{p}_{2}} * 6_{\mathrm{p}_{3}}=\frac{4 !}{2 !} * \frac{6 !}{3 !}=1440

Question:2

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!]

Answer:

Alphabetical order of letters of word RACHIT is A, C, H, I, R,T .
Number of words starting with A=5!
Number of words starting with C=5!
Number of words starting with H=5!
Number of words starting with I=5!
For R, the first word is RACHIT
So, rank of words=4*5!+1=481

Question:3

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.

Answer:

Given :Candidates acn not attempt more than 5 questions from either group and can only attempt
a minimum of 2 questions from either group.
Possible number of questions attempted from each group are :
a) 5 from Group I and 2 from Group II
b) 4 from Group I and 3 from Group II
c) 3 from Group I and 4 from Group II
d) 2 from Group I and 5 from Group II
Total ways= 2\left [^6C_5*^6C_2+^6C_4*^6C_3 \right ]=2*390=780

Question:4

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 =^{18}C_2 -^5C_2 + 1 ]

Answer:

Number of straight lines formed by joining 18 points taking 2 at a time= 18C_2
Number of straight lines formed by joining 5 collinear points taking 2 at a time=5C_2
5 collinear points when joined pairwise give only one line.
Required number of straight lines= 18C_2-5C_2+1=153-10+1=144

Question:5

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?

Answer:

Number of persons to be chosen out of 8=6
CASE I:When A is chosen, B must be chosen
Number of ways of selecting 4 more persons from remaining 6 persons= ^{8-2}C_{6-2}=6C_4=15
CASE II:When A is not chosen
Number of ways of selecting 6 persons from remaining 7 persons= 7C_6=7
Total ways=15+7=22

Question:6

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} C_4.]

Answer:

Out of 12 persons, a chairperson can be selected in ^{12}C_1=12 ways.
The remaining 4 persons can be selected out of 11 persons in ^{11}C_4=330 ways
Total number of committe=12*330=3960

Question:7

How many automobile license plates can be made if each plate contains two different letters followed by three different digits?

Answer:

Each plate contains 2 different letters followed by 3 different digits
Arrangement of 26 letters taken 2 at a time=^{26}P_2=26*25=650
Arrangementof 10 digits taken 3 at a time= ^{10}P_3=10*9*8=720
Total number of license plates=650*720=468000

Question:8

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.

Answer:

2 black balls can be selected from 5 black balls in ^5C_2ways
3 red balls can be selected from 6 red balls in ^6C_3 ways
Total number of ways= ^5C_2* ^6C_3=10*20=200

Question:9

Find the number of permutations of n distinct things taken r together, in which 3 particular things must occur together.

Answer:

Combination of n things taken r at a time in which 3 things always occur together=^ {n-3}C_{r-3}
3 things taken together is considered 1 group Number of arrangement of 3 things=3!
Now, number of arrangement of r-2 objects=(r-2)!
Total number of arrangements=^{n-3}C_{r-3}*(r-2)!*3!

Question:10

Find the number of different words that can be formed from the letters of the word ‘TRIANGLE’ so that no vowels are together.

Answer:

Consonants in TRIANGLE are T, R, N, G, L . Vowels are I, A, E
Five consonants can be arranged in 5! ways XCXCXCXCXCX
Arrangements of consonants as C above creates 6 gaps marked as X
Now, 3 vowels can be arranged in any three of these 6 gaps in ^6P_3 ways
So, total number of arrangements=5!*^6P_3=120*120=14400

Question:11

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.

Answer:

A number is divisible by 5, if at the unit place of the number there is 0 or 5. So, unit digit can be filled in 2 ways. Since, we have to form 4-digit numbers greater than 6000 and less than 7000. ‘6’ can fill the thousandth place only.

The hundredth and tenth place can be filled together in 8*7 =56 ways. So, total number of ways = 56*2 =112

Question:12

There are 10 persons named P_1,P_2,P_3, ... P_{10}. 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 =7C_4\times 5!]

Answer:

Given that P_1,P_2,P_3, ... P_{10}
persons out of which 5 are to be arranged but P_1must occur
whereas P4and P5 never occur. So, now we only have to select 4 out of 7 persons
Number of selections=^7C_4=35
Number of the arrangement of 5 persons=35*5!=35*120=4200

Question:13

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].

Answer:

The hall can be illuminated if atleast one lamp is switched out of 10 lamps Total number of ways
=^{10}C_{1}+^{10}C_{2}+\cdots \cdot \ldots \ldots+^{10}C_{10}=2^{10}-1=1024-1=1023

Question:14

A box contains two white, three black and four red balls. In how many ways can three balls be drawn from the box , if at least one black ball is to be included in the draw.
[Hint: Required number of ways ={ }^{3} \mathrm{C}_{1} \times{ }^{6} \mathrm{C}_{2}+{ }^{3} \mathrm{C}_{2} \mathrm{x}^{6} \mathrm{C}_{1}+{ }^{3} \mathrm{C}_{3}]

Answer:

Out of 2 white, 3 black and 4 red balls we have to draw 3 balls such that atleast one black is included.
The possibilities are:
a.1 black ball and 2 others b.2 black balls and 1 other (c)3 black balls
Number of selections = 3_{C_{1}} * 6_{C_{2}}+3_{C_{2}} * 6_{C_{1}}+3_{C_{3}} * 6_{C_{0}}
=3*15+3*6+1=45+18+1=64

Question:15

If ^nC_{r - 1} = 36, ^nC_{r } = 84and ^nC_{r +1} = 126, then find ^rC_{2}.
[Hint: From equation \frac{{ }^{n} C_{r}}{{ }^{n} C_{r+1}} \text { and } \frac{{ }^{n} C_{r}}{{ }^{n} C_{r-1}} using to find the value of r.]

Answer:

\\\frac{n_{C_{r}}}{n_{C_{r-1}}}=\frac{n-r+1}{r}=\frac{84}{36}=\frac{7}{3} \\\\ 3 n-3 r+3=7 r \\ 10 r-3 n=3 \ldots \ldots . .(i ) \\\\ \frac{n_{C_{r+1}}}{n_{C_{r}}}=\frac{n-(r+1)+1}{r+1}=\frac{126}{84} \\ \frac{n-r}{r+1}=\frac{3}{2} \\ 2 n-2 r=3 r+3 \\ 2 n-5 r=3 \ldots \ldots . .(i i)
Equation (i) + 2 Equation(ii)
10r - 3n +4n - 10r = 3 + 6
n = 9
r =3
^rC_2 = ^3C_2 = 3

Question:16

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.]

Answer:

Since, repetition is not allowed a maximum of 5- digit numbers can be formed with digits 3, 5,7, 8 and 9.

Since, all the 5-digit numbers are greater than 7000 the number of 5-digitintegers= 5*4*3*2*1=120

A 4-digit number is greater than 7000 if thousandth place has any of the 7,8 and 9. The remaining 3 places can be filled from remaining 4 digits in ^4P_3ways

So, total number of 4-digit integers = 3*^4P_3=3*4*3*2 =72

Total number of integers=120+72 =192

Question:17

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?

Answer:

If no two lines are parallel, all lines are intersecting, and no three lines are concurrent.

Two straight lines create one point of intersection
Number of points of intersection = Number of combinations of 20 straight lines taken two at a time
=^{20}{C_{2}}=\frac{20 !}{2 ! 18 !}=20 * \frac{19}{2 * 1}=190

Question:18

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 six digits distinct?

Answer:

If first two digit is 41, the remaining 4 digits can be arranged in ^8P_4 ways= 8*7*6*5 =1680

In addition, if first two digits are 42, 46, 62 or 64 the remaining 4 can be arranged in ^8P_4 ways=8*7*6*5 =1680 ways.

So, total number of telephone numbers having all 6 digits different = 5*1680 =8400

Question:19

In an examination, 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.

Answer:

Since, 2 questions are compulsory, the other 2 can be selected from the remaining 3 questions in ^3C_2 ways = 3 ways

Question:20

A convex polygon has 44 diagonals. Find the number of its sides. [Hint: Polygon of n sides has (^nC_2 - n) number of diagonals.]

Answer:

Let the convex polygon have n sides.
Number of diagonals = number of ways of selecting two vertices- number of sides
\\n_{c_{2}}-n=44 \\\\ \frac{n !}{2 !(n-2) !}-n=44 \\\\ \frac{n(n-1)}{2}-n=44 \\\\ n(n-1)-2 n=88 \\\\ n^{2}-3 n-88=0 \\\\ (n-11)(n+8) \\\\ n=11

Question:21

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?

Answer:

Since, 18 mice were placed equally in three groups. Each group had 6 mice
Number of ways of forming 3 groups each with 6 mice= ^{18}C_6*^{12}C_6*^{6}C_6=\frac{18!}{6!6!6!}
But this includes the order in which the three groups are formed i.e. 3! which should not be
counted.
So, actual number of ways=\frac{18!}{6!6!6!3!}

Question:22

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.

Answer:

Total number of marbles=6W+5R=11

a.If they can be of any color we have to select 4 out of 11 marbles= 11C_4
b.2 white marbles can be selected in ^6C_2ways and 2 red marbles in ^5C_2.
Total number of ways= ^6C_2*^5C_2=15*10=150
c. If they all are of same color 4 white out of 6 marbles can be selected in ^6C_4ways and
4 red out of 5 can be selected in ^5C_4 ways
Required number of ways=^6C_4+^5C_4=15+5=20

Question:23

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?

Answer:

Number of ways to select a team of 11 players out of 16= ^{16}C_{11}
a. If two particular players are include then 9 more can be selected from remaining 14 players= ^{14}C_{9}
b. If two particular players are excluded then all 11 players can be selected from remaining 14
players in ^{14}C_{11}

Question:24

A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and at least 5 from Class X II. If there are 20 students in each of these classes, in how many ways can the team be constituted?

Answer:

We have to select atleast 5 students from each class out of 20 students in each class . So, we
can either select 5 students from class XI and 6 from class XII or vice versa
Total number of ways= 20 c_{5} * 20 c_{6}+20 c_{6} * 20 c_{5}=2\left(20 c_{6} \cdot 20_{c_{5}}\right)

Question:25

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.

Answer:

i.Team having no girls= ^7C_5=\frac{7*6}{2!}=21
ii.Team having atleast one boy and one girl= ^7C_1*^4C_4+^7C_2*^4C_3+^7C_3*^4C_2+^7C_4*^4C_1=7*1+21*4+35*6+35*4=441
iii.Team having atleast 3 girls = ^4C_3*^7C_2+^4C_4*^7C_1=4*21+7=91

Question:26

If ^nC_{12} = ^nC_8, then n is equal to
A. 20
B. 12
C. 6
D. 30

Answer:

The answer is the option a)
We have, ^nC_{12} = ^nC_8
\\ ^nC_{n-12} = ^nc_8 \\ n-12=8 \\ n=20

Question:27

The number of possible outcomes when a coin is tossed 6 times is
A. 36
B. 64
C. 12
D. 32

Answer:

The answer is the option (b) number of outcomes when a coin is tossed= 2 (head or tail)

Total possible outcomes when a coin tossed 6 times = 2*2*2*2*2*2 =64

Question:28

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

Answer:

The answer is the option (c)
We have to form 4-digit numbers using 2,3,4 and 7.
Required number of ways = ^4P_4=4!=24

Question:29

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

Answer:

The answer is the option (b)
If we fix 3 at the unit place, the remaining places can be filled in 3! Ways.
Thus ‘3’ appears in unit place in 3! Times.
Similarly for each of the digits 4,5 and 6. So, sum of digits in unit place = 3!(3+4+5+6) =18*6 =108

Question:30

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

Answer:

The answer is the option (c)

With 4 vowels and 5 consonants we have to form words with 2 vowels and 3 consonants.

Total number of ways = ^4C_2*^ 5C_3*5!=6*10*120=7200

Question:31

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.]

Answer:

The answer is the option (a).
A number is divisible by 3 if the sum of its digit is divisible by 3. So, to form a number of 5-digit which is divisible by 3 we can remove either 0 or 3. If digits 1,2, 3, 4, 5 are used then the number of required numbers = 5!
If digits 0,1,2,4,5 are used then first place from left can be filled in 4 ways and remaining 4 places can be filled in 4! Ways. So, 4*4! Ways.
Total numbers = 120+96 =216

Question:32

Everybody in a room shakes hands with everybody else. The total number of handshakes is 66. The total number of persons in the room is
A. 11
B. 12
C. 13
D. 14

Answer:

The answer is the option b.

\\ \text { Number of hand shakes }=n_{C_{2}}=\frac{n !}{(n-2) ! 2 !}=\frac{n(n-1)}{2}=66 \\ n(n-1)=132 \\ n^{2}-n=132=0 \\ (n-12)(n+11)=0 \\ n=12

Question:33

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

Answer:

The answer is the option (d).
Number of ways of selecting 3 points from given 12 points=^{12}C_3
But 3 points from given 7 collinear points does not form triangle.
Number of ways of selecting 3 points from seven collinear points= ^{7}C_3
Total number of triangles=^{12}C_3-^{7}C_3=220-35=185

Question:34

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

Answer:

The answer is the option (b).

We need to select a pair of line from a set of 4 lines and another pair from another set of 3 lines to form a parallelogram

Required number = ^4C_2*^3C_2=6*3=18

Question:35

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.^{16}C_{11}
B.^{16}C_{5}
C.^{16}C_{9}
D.^{20}C_{9}

Answer:

The answer is the option (c).
We have to select 11 players out of 22.
After excluding particular 4 of them, 18 are available. Amongst these 18, 2 particular players are always included. Thus, we have to select 9 more players from the remaining 16 players in ^{16}C_{9} ways.

Question:36

The number of 5-digit telephone numbers having at least one of their digits repeated is
A. 90,000
B. 10,000
C. 30,240
D. 69,760

Answer:

The answer is the option (d).
Total number of telephone numbers when there is no restriction = 10^5
Number of telephone numbers with all digits different = ^{10}P_5
Required number of ways=10^5-^{10}P_5 =100000-10*9*8*7*6
=100000-30240=69760

Question:37

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

Answer:

The answer is the option (a).

To satisfy the condition mentioned in the question we can either select 2 men and 4 women or 3 men and 6 women =

^4C_2*^6C_4+^4C_3*^6C_6=6*15+4*1=94

Question:38

The total number of 9-digit numbers which have all different digits is
A. 10!
B. 9!
C. 9 × 9!
D. 10×10!

Answer:

The answer is the option (c).
We have to form 9-digit number, which has all different digits.
First digit from the left can be filled in 9 ways excluding 0. After this, nine digits are left including ‘0’. Remaining 8 places can be filled with these nine digits in ^9P_8 ways. So, total numbers= 9* ^9P_8 =9*9!

Question:39

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.^4C_4 \times ^3C_3

Answer:

The answer is the option (b)
Vowels in ARTICLE are A,I,E and consonants are R,T,C,L . Vowels occupy 3 even places in 3! Ways and in remaining 4 places 4 consonants can be arranged in 4! Ways
Total number of words = 3!* 4! =6*24 =144

Question:40

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 2^5, 2^4 \text{and } 2^3, respectively.]

Answer:

The answer is the option (b).
At least one green dye can be chosen in ^5C_1+^5C_2+^5C_3+^5C_4+^5C_5=2^5-1ways
At least one blue dye can be chosen in ^4C_1+^4C_2+^4C_3+^4C_4=2^4-1ways
Any number of red dyes can be chosen in ^3C_0+^3C_1+^3C_2+^3C_3=2^3ways
So, total number of selection=\left (2^5-1 \right )*\left (2^4-1 \right )*\left (2^3 \right )=3720

Question:41

Fill in the Blanks If ^nP_r = 840, ^nC_r = 35, then r = ______.

Answer:

\\^n{P_{r}}=840 \text { and } ^n{C_{r}}=35 \\ ^n{P_{r}}=^n{C_{r}} * r ! \\ 840=35 * r ! \\ r !=24\: \: , r=4

Question:42

Fill in the Blanks ^{15}C_8 + ^{15}C_9 - ^{15}C_6 - ^{15}C_7 =______.

Answer:

=15 c_{8}+15 c_{9}-15 c_{15-6}-15 c_{15-7}=15 c_{8}+15 c_{9}-15 c_{8}-15 c_{9}=0

Question:43

Fill in the Blanks
The number of permutations of n different objects, taken r at a line, when repetitions are allowed, is ______.
Answer:

n*n*n...r\: times = n^r

Question:44

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 is \frac{6!}{2!} and number of ways of arranging vowels =^7P_6\times \frac{1}{3!}\times \frac{1}{2!}

Answer:

. Vowels in INTERMEDIATE are I, E, E, I, A, E and consonants N, T, R, M, D, T
First six consonants are arranged in \frac{6!}{2!}ways.
In the 7 gaps, six vowels are arranged in ^7C_6*\frac{6!}{2!3!} ways
Total number of words \frac{6!}{2!}*^7C_6*\frac{6!}{2!3!}=360*7*60=151200

Question:45

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 ______.

Answer:

Possible selection can be either '2red and 1 other' or ' 3 red'.
Required number of ways= ^5C_2*^7C_1+^5C_3=10*7+10=80

Question:46

Fill in the Blanks
The number of six-digit numbers, all digits of which are odd is ______.

Answer:

Odd digits are 1, 3, 5, 7, 9

The required number of numbers = 5*5*5*5*5*5 = 5^6

Question:47

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 ______.

Answer:

. Let the number of teams participating in the championship be n.

It is given that every 2 teams played one match with another.

Total match= ^nC_2=153

\frac{n\left ( n-1 \right )}{2}=153

n^2-n-306=0
\\ \left (n-18 \right )\left (n+17 \right )=0\\ n=18

Question:48

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 ______

Answer:

. Number of ways of arrangement of six+signs is 1.
In the 7 gaps created, 4 are to be chosen for '-' signs in ^7C_4 ways.
These signs can only be arranged in 1 way as they are identical.
Total number of ways =1* ^7C_4*1=35 ways

Question:49

Fill in the Blanks A committee of 6 is to be chosen from 10 men and 7 women so as to contain at least 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 = ^{10}C_3 \times ^7C_3 + ^{10}C_4 \times ^7C_2.
For 2 particular women to be always there: the number of ways = ^{10}C_4 + ^{10}C_3 \times^5C_1.
The total number of committees when two particular women are never together = Total – together.]

Answer:

Number of ways of forming a committee of 6 persons containing atleast 3 men and 2 women

= ^{10}C_3* ^7C_3+ ^{10}C_4* ^7C_2=120*35+210*21=8610

Total number of committee when two particular women are never together

=Total number of ways-Number of ways when 2 particular women are together

= \left (^{10}C_3* ^7C_3+ ^{10}C_4* ^7C_2 \right )- \left (^{10}C_4+ ^{10}C_3* ^5C_1 \right )

= 4200+4410-(210+600)=7800

Question:50

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 ______.
Answer:

Required number of ways such that 3 balls are drawn and atleast one black ball is included
in the draw= ^3C_1*^ 6C_2+ ^3C_2* ^6C_1+ ^3C_3
=3*15+3*6+1=45+18+1=64

Question:51

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 ^{12}C_2 - ^{5}C_2.

Answer:

False. Required number of lines= ^{12}C_2 - ^{5}C_2+1

Question:52

State whether the statements in True or False.
Three letters can be posted in five letterboxes in 3^5 ways.

Answer:

False.

Total number of ways of posting 3 letters = 5*5*5 = 125

Question:53

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-m}P_{r-m} \times ^rP_m.

Answer:

False. In arrangement of n things taken r at a time in which m occur together, first we
select \left ( r-m \right ) objects from \left (n-m \right ) objects in ^{n-m}C_{r-m} ways.
If we consider these m things as 1 group number of objects excluding these m objects=\left ( r-m \right )
Now, we have to arrange\left ( r-m+1 \right ) objects
Number of arrangements=\left ( r-m+1 \right )!
Also, m objects which are considered as 1 group can be arranged in m!ways
Required number of arrangements= ^{n-m}C_{r-m}*\left (r-m+1! \right )*m!

Question:54

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 3^{12} ways.

Answer:

True. In each stall any oneof the three animals can be shipped
So, total number of ways of loading=3*3*3*...12 \: \: times=3^{12}

Question:55

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 2^n-1.

Answer:

True.
If some or all objects taken at a time, then the number of combinations would be
n_{C_{2}}+n_{C_{2}}+\cdots \ldots+n_{C_{n}}=2^{n}-1

Question:56

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.

Answer:

False.

Number of ways of selecting any number of objects from n given identical objects is 1. Now, selecting 0 or more red ball from 4 identical red balls= 1+1+1+1+1=5

Selecting at least 1 black ball from 5 identical black balls =1+1+1+1+1=5

So, total ways=5*5 =25

Question:57

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 arrangements can be made is \frac{11 !}{5 ! 6 !}(9 !)(9 !)
[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.]

Answer:

True
Let the two sides of table be A and B with 9 seats each.
Let on side A 4 particular guests and on side B 3 particular guests be seated.
Now, for side A, 5 more guests can be selected can be selected from remaining 11 guests in ^{11}C_5
ways.
Also, on each side nine guests can be arranged in 9!ways.
So, total number of ways of arrangement= 11C_5*9!*9!=\frac{11 !}{5 ! 6 !}(9 !)(9 !)

Question:58

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.

Answer:

False.
A candidate can attempt questions in either of the following ways :
a.2 from group A and 5 from group B
b.3 from group A and 4 from group B
c.4 from group A and 3 from group B
d.5 from group A and 2 from group B
Number of ways of attempting 7 questions
\\=6_{C_{2}} * 6_{C_{5}}+6_{C_{3}} * 6_{C_{4}}+6_{C_{4}} * 6_{C_{3}}+6_{C_{5}} * 6_{C_{2}}\\ =2\left(6 c_{2} * 6 c_{5}+6 c_{3} * 6_{c_{4}}\right)=2(15 * 6+20 * 15)=2(90+300)=2 * 390=780

Question:59

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 ^5C_3 \times^{ 22}C_9.

Answer:

True. We can select 3 scheduled caste candidate out of 5 in ^5C_3ways.
And we can select 9 other candidates out of 22 in ^{22}C_9ways.
Total number of selections=^5C_3* ^{22}C_9

Question:60

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:
\begin{array}{|l|l|l|l|} \hline \text { C }_{\mathbf{1}} & & \mathbf{C}_{2} & \\ \hline \text { (a) } & \begin{array}{l} \text { One book of each } \\ \text { subject: } \end{array} & \text { (i) } & 3968 \\ \hline \text { (b) } & \begin{array}{l} \text { At least one book } \\ \text { of each subject: } \end{array} & \text { (ii) } & 60 \\ \hline \text { (c) } & \begin{array}{l} \text { At least one book } \\ \text { of English: } \end{array} & \text { (iii) } & 3255 \\ \hline \end{array}

Answer:

Number of book of mathematics=3
Number of book of physics=4
Number of book of English=5
a.One book of each subject= ^3C_1*^4C_1*^5C_1=60
b.At least one book of each subject=\left (2^3-1 \right )\left (2^3-1 \right )\left (2^5-1 \right )=3225
c. Adding one book of English=\left (2^5-1 \right ).2^7=3968
So, a-(ii), b-(iii), c-(i)

Question:61

Five boys and five girls form a line. Find the number of ways of making the seating arrangement under the following condition:
\begin{array}{|l|l|l|l|} \hline \mathbf{C}_{\mathbf{1}} & & \mathbf{C}_{2} & \\ \hline \text { (a) } & \begin{array}{l} \text { Boys and girls } \\ \text { alternate: } \end{array} & \text { (i) } & \begin{array}{l} \text { 5! } \times \\ \text { 6! } \end{array} \\ \hline \text { (b) } & \begin{array}{l} \text { No two girls sit } \\ \text { together : } \end{array} & \text { (ii) } & \begin{array}{l} 10 ! \\ -5 \\ !6 ! \end{array} \\ \hline \text { (c) } & \begin{array}{l} \text { All the girls sit } \\ \text { together } \end{array} & \text { (iii) } & \begin{array}{l} (5 !)^{2} \\ + \\ (5 !)^{2} \end{array} \\ \hline \text { (d) } & \begin{array}{l} \text { All the girls are } \\ \text { never together : } \end{array} & \text { (iv) } & \begin{array}{l} 2 ! \\ 5 ! \\ 5 ! \end{array} \\ \hline \end{array}

Answer:

Given that number of boys=5 and number of girls=5
a.Boys and girls alternate=\left (5!. 5! \right )+\left (5!. 5! \right )=\left (5! \right )^2+\left (5! \right )^2
b.No two girls sit together=5!. 6!
c.All the girls sit together=2! . 5!. 5!
d .All the girls are never together :
Total number of boys and girls=5+5=10
Number of ways=10!-5!6!

Question:62

There are 10 professors and 20 lecturers out of whom a committee of 2 professors and 3 lecturers is to be formed. Find:

C_1

C_2

a)In how many ways committee:can be formed

i)^{10}C_2\times ^{19}C_3

b)In how many ways a particular: professor is included

ii)^{10}C_2\times ^{19}C_2

c)In how many ways a particular: lecturer is included

iii) ^{9}C_1\times ^{20}C_3

d)n how many ways a particular: lecturer is excluded

iv)^{10}C_2\times ^{20}C_3

Answer:

Given that number of professors=10and number of lecturers=20
a.In how many ways committee can be formed= ^{10}C_2\times ^{20}C_3
b.In how many ways a particular professor is included= ^{9}C_1\times ^{20}C_3
c.In how many ways a aprticular lecturer is included= ^{10}C_2\times ^{19}C_2
d.In how many ways a particular lecturer is excluded= ^{10}C_2\times ^{19}C_3
So a-iv, b-iii, c-ii, d-i

Question:63

Using the digits 1, 2, 3, 4, 5, 6, 7, a number of 4 different digits is formed. Find
\begin{array}{|l|l|l|l|} \hline \mathbf{C}_{\mathbf{1}} & & \mathbf{C}_{\mathbf{2}} & \\ \hline \text { (a) } & \begin{array}{l} \text { how many } \\ \text { numbers are } \\ \text { formed? } \end{array} & \text { (i) } & 840 \\ \hline \text { (b) } & \begin{array}{l} \text { how many } \\ \text { numbers are } \\ \text { exactly divisible } \\ \text { by 2? } \end{array} & \text { (ii) } & 200 \\ \hline \text { (c) } & \begin{array}{l} \text { how many } \\ \text { numbers are } \\ \text { exactly divisible } \\ \text { by 25? } \end{array} & \text { (iii) }&360 \\ \hline \text { (d) } & \begin{array}{l} \text { how many of } \\ \text { these are exactly } \\ \text { divisible by 4? } \end{array} & \text { (iiv) } & 40 \\ \hline \end{array}

Answer:

Given :Total number of digit to be formed=4
a. How many numbers are formed=^7P_4=840
b. How many numbers are exactly divisible by 2?
Number ending with 2= ^6P_3=120
Number ending with 4 = ^6P_3=120
Numbers ending with 6 = ^6P_3=120
So, total numbers=120+120+120=360
c. Numbers exactly divisible by 25=40
d. The numbers which have last 2 digit divisible by 4 are 12, 16 , 24 , 32, 36 , 44 ,52, 56, 64, 72, 76 So = ^{11}P_2*^ 5P_2=200

Question:64

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
\begin{array}{|l|l|l|l|} \hline \mathbf{C}_{\mathbf{1}} & & \mathbf{C}_{2} & \\ \hline \text { (a) } & \begin{array}{l} \text { 4 letters are used } \\ \text { at a time } \end{array} & \text { (i) } & 720 \\ \hline \text { (b) } & \begin{array}{l} \text { All letters are } \\ \text { used at a time } \end{array} & \text { (ii) } & 240 \\ \hline \text { (c) } & \begin{array}{l} \text { All letters are } \\ \text { used but the first } \\ \text { is a vowel } \end{array} & \text { (iii) } & 360 \\ \hline \end{array}

Answer:

Given that Total number of letter of MONDAY=6
Total vowel=2
a. When 4 letters are used at a time= ^6P_4=360
b. All the letters are used at a time = ^6P_6=720
c. All letters are used but the first is a vowel=2. 5!

Important Notes From NCERT Exemplar Class 11 Maths Solutions Chapter 7 Permutations and Combinations

The letter ‘n’ is important when it comes to formulas, where any number to be found said to be it. It is the first natural number and the product procured by ‘n’ is denoted as ‘n!’, known as factorial notations.’ Through NCERT Exemplar Class 11 Maths Chapter 7 solutions, students will learn about the formula derivation, along with the letter used and its role.

Combinations could be made in any manner without any restrictions, all answers will be correct. On the other hand, in permutations case, they need to be arranged properly and in a correct order to get definite results. All of this is explained in the NCERT Exemplar solutions for Class 11 Maths chapter 7. By using NCERT Exemplar Class 11 Maths Chapter 7 solutions PDF Download, students can make learning even more convenient. Here, this guide would be easy for students to solve their issues and easily understand this topic.

Main Subtopics in NCERT Exemplar Class 11 Maths Chapter 7 Solutions

  • Introduction
  • Fundamental Principle of Counting
  • Permutations
  • Permutations when all the objects are distinct
  • Factorial notation
  • Derivation of the formula for nPr
  • Permutations when all the objects are not distinct
  • Combinations

What will the students learn from NCERT Exemplar Class 11 Maths Chapter 7 Solutions?

NCERT Exemplar Class 11 Chapter 7 solutions deals with a subcategory of probability. Students here will know about the difference between permutations and combinations, which has a fine line of separation. Not to be confused, this solution will help students identify the arrangements and pairing portions.

Whenever probability will come in picture, whether basic or advanced both of these matters come in their role-play and play their part. Resolving issues using these formulae will be helpful.

As per NCERT Exemplar Class 11 Maths Chapter 7 solutions, Questions on compound events could be easily solved using these techniques. In combination and permutation, the issue is resolved by knowing that one factor, about arranging the objects correctly in the right order is main, and the pairing could be done in any manner.

In NCERT Exemplar Class 11 Maths Solutions Chapter 7 the students will learn about probability solving questions, along with the number of combinations that could be made from it, that could be very useful for future uses. Easy examples help kids do it better and cleverly.

NCERT Solutions for Class 11 Mathematics Chapters

Important Topics in NCERT Exemplar Class 11 Maths Chapter 7 Solutions Permutations and Combinations

· Class 11 Maths NCERT Exemplar solutions Chapter 7 has detailed that Principles of counting, permutations, its formula and other subtopics and combinations are important topics which students should pay extra attention to.

· As per NCERT Exemplar Class 11 Maths solutions Chapter 7, Permutations and Combinations are said to be the ‘fundamentals principle of counting’- leading to arranging and combining objects. It is about how the students will understand the different situations occurring at different periods, and what could be the possible answers derived to those questions.

Check Chapter-Wise NCERT Solutions of Book

NCERT Exemplar Class 11 Solutions

Read more NCERT Solution subject wise -

Also, read NCERT Notes subject wise -

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Frequently Asked Questions (FAQs)

1. Is this chapter crucial for higher education?

Yes, permutations and combinations are one of the most important chapters of maths that will create a base for board exams and entrance exams. 

2. How many questions are solved in these solutions?

In total, 42 questions are solved from four exercises and miscellaneous exercises from the NCERT chapter in NCERT Exemplar Class 11 Maths Chapter 7 Solutions. 

3. Are these solutions as per the marking scheme?

Yes, these solutions are designed stepwise as per the CBSE pattern so that one can see how the marks are divided among each step. 

4. Are these solutions accessible offline?

Yes, these NCERT Exemplar Class 11 Maths Solutions Chapter 7 are downloadable in PDF format so that one can refer to solutions while solving without being online.

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