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

Chapter 7 Permutations and Combinations Exercise 7.4 NCERT Solutions for Class 11 Maths will help you in boosting your marks. Class 11 Maths NCERT Solutions is updated as per the latest syllabus of CBSE so you can easily make your homework. You can also frame your own answers by taking help from these detailed NCERT Solutions.

1. If

We have

2. Determine n if

^{n}C_{8}=^{n}C_{2}. Find^{n}C_{2}.**Answer**We have

^{n}C_{r}=^{n}C_{n – r}^{n}C_{2}=^{n}C_{n – 2}^{n}C_{8 }=^{n}C_{n – 2 }=> n – 2 = 8 or n = 10^{n}C_{2}=^{10}C_{2}= (10 × 9)/(1 × 2) = 452. Determine n if

(i)

^{2n}C_{3}:^{n}C_{2}= 12: 1
(ii)

^{2n}C_{3}:^{n}C_{3}= 11: 1**Answer****(i)**

^{2n}C

_{3}:

^{n}C

_{2}= 12: 1

=> (2n(2n – 1)(2n – 2))/1.2.3 ÷ (n(n – 1))/1.2 = 12/1

[

^{n}C_{r}= (n(n – 1)……(n – r + 1))/(1.2.3 ……..n)]
Or (2n(2n – 1)2(n – 1))/6 × 2/(n(n – 1)) = 12/1

Or (4n(2n -1)(n – 1))/3 × 1/(n(n – 1)) = 12

2n – 1 = 9, 2n = 10 or n = 5

**(ii)**

^{2n}C

_{3}:

^{n}C

_{3}= 11: 1

Or (2n(2n – 1)(2n – 2))/1.2.3 ÷ (n(n – 1)(n – 2))/1.2.3 = 11/1

Or (4n(n -1)(2n – 1))/6 × 6/(n(n -1)(n – 2)) = 11/1

Or (4(2n -1))/(n -2) = 11

4(2n – 1) = 11(n – 2) or 8n – 4 = 11n – 22

Or 3n = 18 ∴ n = 6

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

We get a chord by joining two points

If P is the number of chords from 21 points, then

P = C(21, 2) = 21!/(2!(21 – 2)!) = 21!/(2!19!)

= (21× 20(19!))/(2 × (19!)) = 21 × 10 = 210 chords

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

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

**Answer**We get a chord by joining two points

If P is the number of chords from 21 points, then

P = C(21, 2) = 21!/(2!(21 – 2)!) = 21!/(2!19!)

= (21× 20(19!))/(2 × (19!)) = 21 × 10 = 210 chords

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

**Answer**
3 boys out of 5 boys can be selected in

^{5}C_{3}ways.
3 girls out of 4 girls can be selected in

^{4}C_{3}ways.
Number of ways in which 3 boys and 3 girls are selected =

^{5}C_{3}×^{4}C_{3}=^{5}C_{2}×^{4}C_{1}
= (5 × 4)/(1 × 2) × 4/1 = 40

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.

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.

**Answer**
The number of ways of selecting 3 red balls out of 6 red balls =

^{6}C_{3}
The number of ways of selecting 3 white balls out of 5 white balls =

^{5}C_{3}
The number of ways of selecting 3 blue balls out of 5 blue balls =

^{5}C_{3}
The number of ways of selecting 3 balls of each colour =

^{6}C_{3 }×^{5}C_{3}×^{5}C_{3}=^{6}C_{3}×^{5}C_{2}×^{5}C_{2}
= ( 6 × 5 × 4)/ ( 1 × 2 ×3) × (5 × 4)/( 1 × 2 ) × ( 5 × 4)/(1 × 2) = 20 × 10 × 10 = 2000

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

One ace will be selected from four aces and four cards will be selected from 52 – 4 = 48 cards

If P is the required number of ways, then P = C(4, 1) × C(48, 4)

= 4!/(1!(4 – 1)!) × 48!/(4!(48 – 4)!)

= (4 × 3!)/(1! × 3!) × ( 48 × 47 × 46 × 45 × 44!)/( 4 × 3 × 2 × 1 × 44!)

= 4 × 2 × 47 × 46 × 45 = 778320 ways

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

Four blowers can be selected from the five blowers and seven players can be selected from 12 players (17 – 5 = 12). If P is the number of ways of selecting the cricket team of eleven, then

P = C(5, 4) × C(12, 7)

= 5! /(4!(5 – 4) !) × 12!/(7!(12 – 7)!)

= (5 × 4!)/(1 × 4!) × (12 × 11 × 10 × 9 × 8 × 7!)/(5! × 7!) = 3960

8. A bag contain 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

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

**Answer**One ace will be selected from four aces and four cards will be selected from 52 – 4 = 48 cards

If P is the required number of ways, then P = C(4, 1) × C(48, 4)

= 4!/(1!(4 – 1)!) × 48!/(4!(48 – 4)!)

= (4 × 3!)/(1! × 3!) × ( 48 × 47 × 46 × 45 × 44!)/( 4 × 3 × 2 × 1 × 44!)

= 4 × 2 × 47 × 46 × 45 = 778320 ways

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

**Answer**Four blowers can be selected from the five blowers and seven players can be selected from 12 players (17 – 5 = 12). If P is the number of ways of selecting the cricket team of eleven, then

P = C(5, 4) × C(12, 7)

= 5! /(4!(5 – 4) !) × 12!/(7!(12 – 7)!)

= (5 × 4!)/(1 × 4!) × (12 × 11 × 10 × 9 × 8 × 7!)/(5! × 7!) = 3960

8. A bag contain 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.

**Answer**
The number of ways in which 2 black balls out of 5 black balls are selected =

^{5}C_{2}
The number of ways in which 3 red balls out of 6 red balls are selected =

^{6}C_{3}
The number of ways in which 2 black and 3 red balls can be selected =

^{5}C_{2}×^{6}C_{3}
= (5 × 4)/(1 × 2) × (6 × 5 × 4)/(1 × 2 × 3) = 10 × 20 = 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?

Out of available nine courses, two are compulsory.

Hence, the student is free to select 3 courses out of 7 remaining courses. If P is the number of ways of selecting 3 courses out of 7 courses, then

P = C(7, 3) = 7!/(3!(7 – 3)!) = 7!/3!4!

= (7 ×6 × 5 × 4!)/(3 × 2 ×1 × 4!) = 7 × 5 = 35 ways

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?

**Answer**Out of available nine courses, two are compulsory.

Hence, the student is free to select 3 courses out of 7 remaining courses. If P is the number of ways of selecting 3 courses out of 7 courses, then

P = C(7, 3) = 7!/(3!(7 – 3)!) = 7!/3!4!

= (7 ×6 × 5 × 4!)/(3 × 2 ×1 × 4!) = 7 × 5 = 35 ways