NCERT Solutions Class 11 Chapter 6 Permutations And Combinations Exercise 6.4 (original) (raw)
Last Updated : 25 Jul, 2024
**Question 1. If **n C 8 = n C 2 , find **n C 2 .
**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) **2n C 3 ****:** **n C 3 **= 12:1 (ii) **2n C 3 : **n C 3 = 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
we have to select 1 Ace card from 4 ace cards
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
- 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
Number of ways to select 4 bowlers from 5=5C4=\frac{5!}{4!1!} =5
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