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

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 6.

The selection of subsets is called a permutation when the order of selection is a factor, while it is called a combination when the order is not a factor. Chapter 7, Permutations and Combinations of Class 11 Maths is categorised under the CBSE Syllabus for 2023-24. 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 practise. The solutions help 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. Also, students can utilise NCERT Class 11 Maths Solutions for board exam preparations.

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

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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.

∴ The number of chords is

⇒ ²¹C₂=

∴ The total number of chords that can be drawn is 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.

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

⇒ ⁵C₃ × ⁴C₃ =

⇒ ⁵C₃ × ⁴C₃ = 10 × 4 = 40

∴ The 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.

∴ The number of ways of selecting 9 balls is ⁶C₃ ×⁵C₃ × ⁵C₃

∴ 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 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 the question, we need to select 1 Ace card out of the 4 Ace cards.

∴ The 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)

∴ The number of ways to select 4 cards from 48 cards is ⁴⁸C₄

∴ The number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination is 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 to bowl, and we can include 4 bowlers in a team of 11.

∴ The number of ways in which bowlers can be selected is ⁵C₄

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

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

∴ The number of ways we can select these players is ¹²C₇

∴ The total number of combinations possible is ⁵C₄ × ¹²C₇

∴ The number of ways we can select a team of 11 players where 4 players are bowlers from 17 players is 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.

The number of ways we can select 2 black balls from 5 black balls is ⁵C₂

The number of ways we can select 3 red balls from 6 red balls is ⁶C₃

The number of ways 2 black and 3 red balls can be selected is ⁵C₂× ⁶C₃

∴ The number of ways in which 2 black and 3 red balls can be selected from 5 black and 6 red balls is 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 needs to select 5 – 2 = 3 courses

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

∴ The number of ways a student selects 5 courses from 9 courses where 2 specific courses are compulsory is 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

Also explore – NCERT Class 11 Solutions