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Class 11 NCERT Solutions- Chapter 7 Permutations And Combinations – Exercise 7.4
  • Last Updated : 28 Dec, 2020

Question 1. If nC8=nC2, find nC2.

Solution

We know that, nCr=nC(n-r)

For the given question, r=8 and n-r=2

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

OR



Using formula (1),

nC8=nC(n-r)

\frac{n!}{8!(n-8)!}=\frac{n!}{2!(n-2)!} \\i.e. \frac{n}{8(n-8)}=\frac{n}{2(n-2)}

8(n-8)=2(n-2) 

4(n-8)=n-2 

4n-32=n-2 

3n=30 

n=10



As n=10, 

10C2 \frac{10!}{2!8!}=\frac{10*9}{2} =45

Question 2. Determine n if (i) 2nC3 : nC3 = 12:1     (ii) 2nC3 : nC3 = 11:1 

Solution:

i) \frac {^{2n}C_3 }{^{n}C_3}=\frac{12}{1} \\\,^{2n}C_3 =12(^{n}C_3) \\\frac{(2n)!}{3!(2n-3)!}=12(\frac{n!}{3!(n-3)!}) \\\frac{(2n)!}{(2n-3)!}=12(\frac{n!}{(n-3)!}) \\\frac{2n(2n-1)(2n-2)(2n-3)!}{(2n-3)!}=12(\frac{n(n-1)(n-2)(n-3)!}{(n-3)!})

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) \frac {^{2n}C_3 }{^{n}C_3}=\frac{11}{1} \\\,^{2n}C_3 =11(^{n}C_3) \\\frac{(2n)!}{3!(2n-3)!}=11(\frac{n!}{3!(n-3)!}) \\\frac{(2n)!}{(2n-3)!}=11(\frac{n!}{(n-3)!}) \\\frac{2n(2n-1)(2n-2)(2n-3)!}{(2n-3)!}=11(\frac{n(n-1)(n-2)(n-3)!}{(n-3)!})

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

= 21C2 =\frac{21!}{2!19!}=\frac{21*20}{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 = 5C3= \frac{5!}{3!2!}  = 10

Number of ways to select 3 girls = 4C3\frac{4!}{3!1!} = 4

Hence, 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= 6C3 \frac{6!}{3!3!}  =20

Number of ways to select 3 balls from 6 red balls= 5C3\frac{5!}{3!2!}  =10

Number of ways to select 3 balls from 6 red balls= 5C3 =\frac{5!}{3!2!}  =10

Number 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= 4C1\frac{4!}{1!3!} = 4

2) Number of ways to select remaining 4 cards

= 48C4 =\frac{48!}{4!44!}=\frac{48*47*46*45}{4*3*2}=194580

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=5C4=\frac{5!}{4!1!} =5

2) Number of ways to select remaining 11-4=7 players from 17-5=12 players

= 12C7 \frac{12!}{7!5!} =792

And, 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= 5C2 =\frac{5!}{2!3!}  =10

Number of ways to select 3 red balls = 6C3\frac{6!}{3!3!}  =20

Hence, 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= 7C3 * 2C2\frac{7!}{3!4!}*\frac{2!}{2!0!}  =35*1 =35

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