# Class 11 NCERT Solutions- Chapter 7 Permutations And Combinations – Exercise 7.4

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

^{n}C

_{8}=

^{n}C

_{2}, find

^{n}C

_{2}

^{.}

**Solution**:

We know that,

^{n}C_{r}=^{n}C_{(n-r)}For the given question, r=8 and n-r=2

Hence, n=r+(n-r)=8+2=10

ORUsing formula (1),

^{n}C_{8}=^{n}C_{(n-r)}8(n-8)=2(n-2)

4(n-8)=n-2

4n-32=n-2

3n=30

n=10

As n=10,

^{10}C_{2 }=

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

^{2n}C

_{3 }:

^{n}C

_{3 }= 12:1 (ii)

^{2n}C

_{3}:

^{n}C

_{3}= 11:1

**Solution**:

i)2n(2n-1)(2n-2)=12n(n-1)(n-2)

(2n-1)2(n-1)=6(n-1)(n-2)

2n-1=3(n-2)

2n-1=3n-6

n=5

ii)2n(2n-1)(2n-2)=11n(n-1)(n-2)

2(2n-1)2(n-1)=11(n-1)(n-2)

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

8n-4=11n-22

3n=18

n=6

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

**Solution**:

Chord of a circle is made by using any two points on a circle. So, we have to select any 2 points from 21 to draw a chord.

Hence, chords that can be drawn through 21 points on a circle

=

^{21}C_{2 }== 210

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

**Solution**:

We have to select 3 boys from 5 boys and 3 girls from 4 girls to make a team.

Number of ways to select 3 boys =

^{5}C_{3}=_{ }= 10Number of ways to select 3 girls =

^{4}C_{3}= = 4Hence, Number of ways to make a required team = 10*4 = 40

**Question 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: **

We have to select 3 balls from 6 red balls, 3 from 5 white balls and 3 from 5 blue balls.

Number of ways to select 3 balls from 6 red balls=

^{6}C_{3 }= =20Number of ways to select 3 balls from 6 red balls=

^{5}C_{3}= =10Number of ways to select 3 balls from 6 red balls=

^{5}C_{3 }= =10Number of ways to select 9 balls in required way=20*10*10=2000

**Question 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**:

We have to select 5 cards from 52 cards. If there is exactly one ace in each combination, then

1) we have to select 1 Ace card from 4 ace cards

2) we have to select 5-1=4 cards from remaining 52-4=48 cards

So, 1) Number of ways to select Ace card=

^{4}C_{1}= = 42) Number of ways to select remaining 4 cards

=

^{48}C_{4 }=And, hence required total number of 5 card combinations=4*194580=778320.

**Question 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**:

We have to select 11 players from 17 players. Among 17 players, 5 are bowlers. So, if there are exactly 4 bowlers to be selected in team of 11 players, then

1) Number of ways to select 4 bowlers from 5=

^{5}C_{4}==52) Number of ways to select remaining 11-4=7 players from 17-5=12 players

=

^{12}C_{7 }= =792And, hence required total number of ways to select a cricket team=792*5=3960

**Question 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**:

We have to select 2 balls from 5 black balls and 3 balls from 6 red balls.

Number of ways to select 2 black balls=

^{5}C_{2 }= =10Number of ways to select 3 red balls =

^{6}C_{3}= =20Hence, Number of ways to make a required team = 10*20=200

**Question 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**:

A student choose 5 courses. Among these 5 courses 2 specific courses are compulsory. Hence, student have to choose 5-2=3 courses from available 9-2=7 courses.

Hence, Number of ways a student can choose a programmer of 5 courses=

^{7}C_{3}*^{2}C_{2}= =35*1 =35