Chapter 7: Permutations and Combinations Exercise 7.4

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

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.

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

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

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.

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.

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.

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.

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