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

### Permutations and Combinations Exercise 7.4 Solutions** **

**1. If ^{n}C_{8} = ^{n}C_{2}, find ^{n}C_{2}.**

**Solution**

It is known that, ^{n}C_{a} = ^{n}C_{b} ⇒ a = b or n = a+ b

Therefore,

^{n}C_{8} = ^{n}C_{2} ⇒ n = 8 + 2 = 10

∴ ^{n}C_{2} = ^{10}C_{2} =

**2. Determine n if (i) ^{2n}C_{3} = ^{n}C_{3} = 12 : 1 (ii) ^{2n}C_{3} = ^{n}C_{3} = 11 : 1 **

**Solution**

(i)

⇒ 2n - 1 = 3(n - 2)

⇒ 2n - 1 = 3n - 6

⇒ 3n - 2n = -1 + 6

⇒ n = 5

(ii)

⇒ 4(2n - 1) = 11(n - 2)

⇒ 8n - 4 = 11n - 22

⇒ 11n - 8n = -4 + 22

⇒ 3n = 18

⇒ n = 6

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

**Solution**

For drawing one chord on a circle, only 2 points are required.

To know the number of chords that can be drawn through the given 21 points on a circle, the number of combinations have to be counted.

Therefore, there will be as many chords as there are combinations of 21 points taken 2 at a time.

Thus, required number of chords = ^{21}C_{2} = 21!/2!(21 -2)! = 21!/2!19! = (21×20)/2 = 210

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

**Solution**

A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.

3 boys can be selected from 5 boys in ^{5}C_{3} ways .

3 girls can be selected from 4 girls in ^{4}C_{3} ways.

Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected = ^{5}C_{3} × ^{4}C_{3} = (5!/3!2!) × (4!/3!1!)

**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**

There are a total of 6 red balls, 5 white balls, and 5 blue balls.

9 balls have to be selected in such a way that each selection consists of 3 balls of each colour.

Here,

3 balls can be selected from 6 red balls in ^{6}C_{3} ways.

3 balls can be selected from 5 white balls in ^{5}C_{3} ways.

3 balls can be selected from 5 blue balls in ^{5}C_{3} ways.

Thus, by multiplication principle, required number of ways of selecting 9 balls

**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**

In a deck of 52 cards, there are 4 aces. A combination of 5 cards have to be made in which there is exactly one ace.

Then, one ace can be selected in ^{4}C_{1} ways and the remaining 4 cards can be selected out of the 48 cards in ^{48}C_{4} ways.

Thus, by multiplication principle, required number of 5 card combinations

**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**

Out of 17 players, 5 players are bowlers.

A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.

4 bowlers can be selected in ^{5}C_{4} ways and the remaining 7 players can be selected out of the 12 players in ^{12}C_{7} ways.

Thus, by multiplication principle, required number of ways of selecting cricket team

**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**

There are 5 black and 6 red balls in the bag.

2 black balls can be selected out of 5 black balls in ^{5}C_{2} ways and 3 red balls can be selected out of 6 red balls in ^{6}C_{3} ways.

Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls

**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**

There are 9 courses available out of which, 2 specific courses are compulsory for every student.

Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be chosen in ^{7}C_{3} ways.

Thus, required number of ways of choosing the programme