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

Question:1 If ${}^n{C}_8{=}^n{C}_2,$ find ${}^n{C}_2$

Answer:

Given : ${}^n{C}_8{=}^n{C}_2,$

We know that ${}^n{C}_a{=}^n{C}_b$ $\Rightarrow a=born=a+b$

${}^n{C}_8{=}^n{C}_2,$

$\Rightarrow n=8+2$

$\Rightarrow n=10$

${}^n{C}_2$${=}^{10}{C}_2$

$=\frac{10!}{(10-2)!2!}$

$=\frac{10!}{8!2!}$

$=\frac{10\times 9\times 8!}{8!2!}$

$=5\times 9=45$

Thus the answer is 45

Question:2(i) Determine n if

${}^{2n}{C}_3{:}^n{C}_3=12:1$

Answer:

Given that : $(i){}^{2n}{C}_3{:}^n{C}_3=12:1$

$\Rightarrow \frac{{}^{2n}{C}_3}{{}^n{C}_3}=\frac{12}{1}$

The ratio can be written as

$\Rightarrow \frac{\frac{2n!}{(2n-3)!3!}}{\frac{n!}{(n-3)!3!}}=\frac{12}{1}$

$\Rightarrow \frac{2n!(n-3)!}{(2n-3)!n!}=\frac{12}{1}$

$\Rightarrow \frac{2n\times (2n-1)\times (2n-2)\times (2n-3)!\times (n-3)!}{(2n-3)!\times n\times (n-1)\times (n-2)\times (n-3)!}=\frac{12}{1}$

$\Rightarrow \frac{2n\times (2n-1)\times (2n-2)}{n\times (n-1)\times (n-2)}=\frac{12}{1}$

$\Rightarrow \frac{2\times (2n-1)\times 2}{(n-2)}=\frac{12}{1}$

$\Rightarrow \frac{4\times (2n-1)}{(n-2)}=\frac{12}{1}$

$\Rightarrow \frac{(2n-1)}{(n-2)}=\frac{3}{1}$

$\Rightarrow 2n-1=3n-6$

$\Rightarrow 6-1=3n-2n$

$\Rightarrow n=5$

Question:2(ii) Determine n if

${}^{2n}{C}_3{:}^n{C}_3=11:1$

Answer:

Given that : $(ii){}^{2n}{C}_3{:}^n{C}_3=11:1$

$\Rightarrow \frac{{}^{2n}{C}_3}{{}^n{C}_3}=\frac{11}{1}$

$\Rightarrow \frac{\frac{2n!}{(2n-3)!3!}}{\frac{n!}{(n-3)!3!}}=\frac{11}{1}$

$\Rightarrow \frac{2n!(n-3)!}{(2n-3)!n!}=\frac{11}{1}$

$\Rightarrow \frac{2n\times (2n-1)\times (2n-2)\times (2n-3)!\times (n-3)!}{(2n-3)!\times n\times (n-1)\times (n-2)\times (n-3)!}=\frac{11}{1}$

$\Rightarrow \frac{2n\times (2n-1)\times (2n-2)}{n\times (n-1)\times (n-2)}=\frac{11}{1}$

$\Rightarrow \frac{2\times (2n-1)\times 2}{(n-2)}=\frac{11}{1}$

$\Rightarrow \frac{4\times (2n-1)}{(n-2)}=\frac{11}{1}$

$\Rightarrow 8n-4=11n-22$

$\Rightarrow 22-4=11n-8n$

$\Rightarrow 3n=18$

$\Rightarrow n=6$

Thus the value of n=6

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

Answer:

To draw chords, 2 points are required on the circle.

To know the number of chords on the circle when points on the circle are 21.

Combinations = Number of chords ${=}^{21}{C}_2$

$=\frac{21!}{(21-2)!2!}$

$=\frac{21!}{19!2!}$

$=\frac{21\times 20}{2}$

$=210$

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

Answer:

A team of 3 boys and 3 girls will 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 boys in ${}^4{C}_3$ ways.

Therefore, by the multiplication principle, the number of ways in which a team of 3 boys and 3 girls can be selected ${=}^5{C}_3{\times }^4{C}_3$

$=\frac{5!}{2!3!}\times \frac{4!}{1!3!}$

$=10\times 4=40$

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 color.

Answer:

There are 6 red balls, 5 white balls, and 5 blue balls.

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

3 balls are selected from 6 red balls in ${}^6{C}_3$.

3 balls are selected from 5 white balls in ${}^5{C}_3$

3 balls are selected from 5 blue balls in ${}^5{C}_3$.

Hence, by the multiplication principle, the number of ways of selecting 9 balls ${=}^6{C}_3{\times }^5{C}_3{\times }^5{C}_3$

$=\frac{6!}{3!3!}\times \frac{5!}{2!3!}\times \frac{5!}{2!3!}$

$=20\times 10\times 10=2000$

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.

Answer:

In a deck, there are 4 aces out of 52 cards.

A combination of 5 cards is to be selected containing exactly one ace.

Then, one ace can be selected in ${}^4{C}_1$ ways, and the other 4 cards can be selected in ${}^{48}{C}_4$ ways.

Hence, using the multiplication principle, required the number of 5 card combination $={}^4{C}_1\times {}^{48}{C}_4$

$=\frac{4!}{1!3!}\times \frac{48!}{44!4!}$

$=4\times \frac{48\times 47\times 46\times 45}{4\times 3\times 2}=778320$

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?

Answer:

Out off, 17 players, 5 are bowlers.

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

4 bowlers can be selected in ${}^5{C}_4$ ways and 7 players can be selected in ${}^{12}{C}_7$ways.

Thus, using multiplication priciple, number of ways of selecting the team $={}^5{C}_4\cdot {}^{12}{C}_7$

$=\frac{5!}{1!4!}\times \frac{12!}{5!7!}$

$=5\times \frac{12\times 11\times 10\times 9\times 8}{5\times 4\times 3\times 2}$

$=3960$

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.

Answer:

A bag contains 5 black and 6 red balls.

2 black balls can be selected in ${}^5{C}_2$ ways and 3 red balls can be selected in ${}^6{C}_3$ ways.

Thus, using multiplication priciple, number of ways of selecting 2 black and 3 red balls ${=}^5{C}_2{.}^6{C}_3$

$=\frac{5!}{2!3!}\times \frac{6!}{3!3!}$

$=10\times 20=200$

Question:9 In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?

Answer:

9 courses are available, and 2 specific courses are compulsory for every student.

Therefore, every student has to select 3 courses out of the remaining 7 courses.

This can be selected in ${}^7{C}_3$ ways.

Thus, using the multiplication principle, the number of ways of selecting courses ${=}^7{C}_3$

$=\frac{7!}{3!4!}$

$=\frac{7\times 6\times 5}{3\times 2}$

$=35$