Class 11 NCERT Mathematics Solutions- Chapter 7 Permutations And Combinations - Exercise 7.4

Chapter 7 of the Class 11 NCERT Mathematics textbook, "Permutations and Combinations," explores fundamental concepts of counting and arrangement. Exercise 7.4 focuses on applying these concepts to solve problems related to permutations and combinations, enhancing students' ability to handle various counting scenarios.

NCERT Solutions for Class 11 - Mathematics - Chapter 7 Permutations and Combinations - Exercise 7.4

This section provides detailed solutions for Exercise 7.4 from Chapter 7 of the Class 11 NCERT Mathematics textbook. The exercise involves solving problems related to permutations and combinations, including arrangements and selections of objects. Solutions are presented step-by-step to help students understand and apply permutation and combination principles effectively.

Permutations and combinations are the initial steps to combinatorial analysis which is a branch of mathematics whose main concern is the calculation of probabilities and arrangements. They assist in acquiring solutions on how to select and organize objects and this is important in different practical situations. Read this article in order to learn more about complete exercise 7. 4 problems of the NCERT Mathematics textbook are demonstrated in detail below to ease your understanding of permutations and combinations.

What is Permutation and Combination?

Permutation and Combination are fundamental concepts in combinatorics, the branch of mathematics dealing with counting, arrangement, and combination of objects. They are used to solve problems related to selecting and arranging objects.

Permutations

Permutation is an arrangement of objects in a specific order. The order of arrangement is important in permutations.

Formula: The number of permutations of n distinct objects taken r at a time is given by:

P(n,r)=\frac{n!}{(n-r)!}

Combinations

Combinations is a selection of objects where the order does not matter.

Formula: The number of combinations of n distinct objects taken r at a time is given by:

C(n, r) = \frac{n!}{r!(n-r)!}

Class 11 NCERT Mathematics Solutions- Exercise 7.4

Question 1. If ⁿC₈=ⁿC₂, find ⁿC₂^.

Solution: 

We know that, ⁿC_r=ⁿC_(n-r)

For the given question, r=8 and n-r=2

Hence, n=r+(n-r)=8+2=10

OR

Using formula (1),

ⁿC₈=ⁿC_(n-r)

\frac{n!}{8!(n-8)!}=\frac{n!}{2!(n-2)!} \\i.e. \frac{n}{8(n-8)}=\frac{n}{2(n-2)}

8(n-8)=2(n-2) 

4(n-8)=n-2 

4n-32=n-2 

3n=30 

n = 10

As n=10, 

¹⁰C₂ = \frac{10!}{2!8!}=\frac{10*9}{2} =45

Question 2. Determine n if (i) ²ⁿC₃ : ⁿC₃ = 12:1     (ii) ²ⁿC₃ : ⁿC₃ = 11:1 

Solution:

i) \frac {^{2n}C_3 }{^{n}C_3}=\frac{12}{1} \\\,^{2n}C_3 =12(^{n}C_3) \\\frac{(2n)!}{3!(2n-3)!}=12(\frac{n!}{3!(n-3)!}) \\\frac{(2n)!}{(2n-3)!}=12(\frac{n!}{(n-3)!}) \\\frac{2n(2n-1)(2n-2)(2n-3)!}{(2n-3)!}=12(\frac{n(n-1)(n-2)(n-3)!}{(n-3)!}) 

2n(2n-1)(2n-2)=12n(n-1)(n-2) 

(2n-1)2(n-1)=6(n-1)(n-2) 

2n-1=3(n-2) 

2n-1=3n-6 

n=5

ii) \frac {^{2n}C_3 }{^{n}C_3}=\frac{11}{1} \\\,^{2n}C_3 =11(^{n}C_3) \\\frac{(2n)!}{3!(2n-3)!}=11(\frac{n!}{3!(n-3)!}) \\\frac{(2n)!}{(2n-3)!}=11(\frac{n!}{(n-3)!}) \\\frac{2n(2n-1)(2n-2)(2n-3)!}{(2n-3)!}=11(\frac{n(n-1)(n-2)(n-3)!}{(n-3)!})

2n(2n-1)(2n-2)=11n(n-1)(n-2) 

2(2n-1)2(n-1)=11(n-1)(n-2) 

4(2n-1)=11(n-2) 

8n-4=11n-22 

3n=18 

n=6

Question 3. How many chords can be drawn through 21 points on a circle?

Solution: 

Chord of a circle is made by using any two points on a circle. So, we have to select any 2 points from 21 to draw a chord. 

Hence, chords that can be drawn through 21 points on a circle

= ²¹C₂ =\frac{21!}{2!19!}=\frac{21*20}{2}  = 210

Question 4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?

Solution: 

We have to select 3 boys from 5 boys and 3 girls from 4 girls to make a team. 

Number of ways to select 3 boys = ⁵C₃= \frac{5!}{3!2!}  = 10

Number of ways to select 3 girls = ⁴C₃ = \frac{4!}{3!1!} = 4

Hence, Number of ways to make a required team = 10*4 = 40 

Question 5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Solution: 

We have to select 3 balls from 6 red balls, 3 from 5 white balls and 3 from 5 blue balls.

Number of ways to select 3 balls from 6 red balls= ⁶C₃ = \frac{6!}{3!3!}  =20

Number of ways to select 3 balls from 6 red balls= ⁵C₃ = \frac{5!}{3!2!}  =10

Number of ways to select 3 balls from 6 red balls= ⁵C₃ =\frac{5!}{3!2!}  =10

Number of ways to select 9 balls in required way=20*10*10=2000

Question 6. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.

Solution:

We have to select 5 cards from 52 cards. If there is exactly one ace in each combination, then 

1) we have to select 1 Ace card from 4 ace cards

2) we have to select 5-1=4 cards from remaining 52-4=48 cards

So, 1) Number of ways to select Ace card= ⁴C₁ = \frac{4!}{1!3!} = 4

2) Number of ways to select remaining 4 cards

= ⁴⁸C₄ =\frac{48!}{4!44!}=\frac{48*47*46*45}{4*3*2}=194580

And, hence required total number of 5 card combinations=4*194580=778320.

Question 7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Solution:

We have to select 11 players from 17 players. Among 17 players, 5 are bowlers. So, if there are exactly 4 bowlers to be selected in team of 11 players, then

1) Number of ways to select 4 bowlers from 5=⁵C₄=\frac{5!}{4!1!} =5

2) Number of ways to select remaining 11-4=7 players from 17-5=12 players

= ¹²C₇ = \frac{12!}{7!5!} =792

And, hence required total number of ways to select a cricket team=792*5=3960

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.

Solution: 

We have to select 2 balls from 5 black balls and 3 balls from 6 red balls.

Number of ways to select 2 black balls= ⁵C₂ =\frac{5!}{2!3!}  =10

Number of ways to select 3 red balls = ⁶C₃ = \frac{6!}{3!3!}  =20

Hence, Number of ways to make a required team = 10*20=200

Question 9. In how many ways can a student choose a programme of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Solution: 

A student choose 5 courses. Among these 5 courses 2 specific courses are compulsory. Hence, student have to choose 5-2=3 courses from available 9-2=7 courses.

Hence, Number of ways a student can choose a programmer of 5 courses= ⁷C₃ × ²C₂ = \frac{7!}{3!4!}*\frac{2!}{2!0!}  =35×1 =35

Read More,

• Chapter 7 Permutations And Combinations – Exercise 7.1
• Chapter 7 Permutations And Combinations – Exercise 7.2
• Chapter 7 Permutations And Combinations – Exercise 7.3

Summary

Permutations and combinations are two critical means in the mechanism of selection and arrangement of objects. Permutations take into account the arrangement of objects whereas in this case the grouping is given importance. These concepts are very essential when dealing with questions on probability, statistics, and other issues arising in day to day life.

Related Articles:

• Class 11 NCERT Mathematics Solutions- Chapter 7 Permutations And Combinations - Exercise 7.3
• Class 11 NCERT Mathematics Solutions- Chapter 7 Permutations And Combinations - Exercise 7.2