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

The selection of subsets is called a permutation when the order of selection is a factor, while it is called a combination when order is not a factor. The Exercise 7.4 of NCERT Solutions for Class 11 Maths Chapter 7- Permutations and Combinations is based on the topic Combination. Students can learn more about the concept by solving the problems given in this exercise.

The NCERT textbook provides plenty of questions for the students to solve and practice. The NCERT Solutions for Class 11 Maths helps the students in understanding the most relevant method of answering a question. This way, the students get a clearer idea of the concept, each time they solve the questions present in the NCERT textbook.

Download PDF of NCERT Solutions for Class 11 Maths Chapter 7- Permutations and Combinations Exercise 7.4

Solutions for Class 11 Maths Chapter 7 – Exercise 7.4

1. If ⁿC₈ = ⁿC₂, find ⁿC_2.

Solution:

2. Determine n if
(i) ²ⁿC_3:ⁿC₃ = 12: 1
(ii) ²ⁿC₃: ⁿC₃ = 11: 1

Solution:

Simplifying and computing

⇒ 4 × (2n – 1) = 12 × (n – 2)

⇒ 8n – 4 = 12n – 24

⇒ 12n – 8n = 24 – 4

⇒ 4n = 20

∴ n = 5

⇒ 11n – 8n = 22 – 4

⇒ 3n = 18

∴ n = 6

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

Solution:

Given 21 points on a circle

We know that we require two points on the circle to draw a chord

∴ Number of chords is are

⇒ ²¹C₂= 

∴ Total number of chords can be drawn are 210

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

Solution:

Given 5 boys and 4 girls are in total 

We can select 3 boys from 5 boys in ⁵C₃ ways

Similarly, we can select 3 boys from 54 girls in ⁴C₃ ways

∴ Number of ways a team of 3 boys and 3 girls can be selected is ⁵C₃ × ⁴C₃

⇒ ⁵C₃ × ⁴C₃ = 

⇒ ⁵C₃ × ⁴C₃ = 10 × 4 = 40

∴ Number of ways a team of 3 boys and 3 girls can be selected is ⁵C₃ × ⁴C₃ = 40 ways

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:

Given 6 red balls, 5 white balls and 5 blue balls

We can select 3 red balls from 6 red balls in ⁶C₃ ways

Similarly, we can select 3 white balls from 5 white balls in ⁵C₃ ways

Similarly, we can select 3 blue balls from 5 blue balls in ⁵C₃ ways

∴ Number of ways of selecting 9 balls is ⁶C₃ ×⁵C₃ × ⁵C₃

∴ 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 is ⁶C₃ ×⁵C₃ × ⁵C₃ = 2000

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:

Given a deck of 52 cards

There are 4 Ace cards in a deck of 52 cards.

According to question, we need to select 1 Ace card out the 4 Ace cards

∴ Number of ways to select 1 Ace from 4 Ace cards is ⁴C₁

⇒ More 4 cards are to be selected now from 48 cards (52 cards – 4 Ace cards)

∴ Number of ways to select 4 cards from 48 cards is ⁴⁸C₄

∴ Number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination 778320.

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:

Given 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers

There are 5 players how bowl, and we can require 4 bowlers in a team of 11

∴ Number of ways in which bowlers can be selected are: ⁵C₄

Now other players left are = 17 – 5(bowlers) = 12

Since we need 11 players in a team and already 4 bowlers are selected, we need to select 7 more players from 12.

∴ Number of ways we can select these players are: ¹²C₇

∴ Total number of combinations possible are: ⁵C₄ × ¹²C₇

∴ Number of ways we can select a team of 11 players where 4 players are bowlers from 17 players are 3960

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:

Given a bag contains 5 black and 6 red balls

Number of ways we can select 2 black balls from 5 black balls are ⁵C₂

Number of ways we can select 3 red balls from 6 red balls are ⁶C₃

Number of ways 2 black and 3 red balls can be selected are ⁵C₂× ⁶C₃

∴ Number of ways in which 2 black and 3 red balls can be selected from 5 black and 6 red balls are 200

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:

Given 9 courses are available and 2 specific courses are compulsory for every student

Here 2 courses are compulsory out of 9 courses, so a student need to select 5 – 2 = 3 courses

∴ Number of ways in which 3 ways can be selected from 9 – 2(compulsory courses) = 7 are ⁷C₃

∴ Number of ways a student selects 5 courses from 9 courses where 2 specific courses are compulsory are: 35

Access other exercise solutions of Class 11 Maths Chapter 7- Permutations and Combinations

Exercise 7.1 Solutions 6 Questions

Exercise 7.2 Solutions 5 Questions

Exercise 7.3 Solutions 11 Questions

Miscellaneous Exercise On Chapter 7 Solutions 11 Questions