NCERT Solutions for Class 11 Maths Chapter 6 Exercise 6.4 - 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 $^nC_a=^nC_b$ $\Rightarrow a=b\, \, \, or\, \, n=a+b$

$^{n}C_{8}=^{n}\; \! \! \! \! C_{2},$

$\Rightarrow n=8+2$

$\Rightarrow n=10$

$^{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 : $(i)\; ^{2n}C_{3}:^{n}C_{3}=12:1$

$\Rightarrow \, \, \frac{^{2n}C_3}{^nC_3}=\frac{12}{1}$

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 : $(ii)\; ^{2n}C_{3}:^{n}C_{3}=11:1$

$\Rightarrow \, \, \frac{^{2n}C_3}{^nC_3}=\frac{11}{1}$

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

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

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

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

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

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

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

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

$\Rightarrow \, \, 3n=18$

$\Rightarrow \, \, n=6$

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 $^5C_3$ ways.

3 girls can be selected from 4 boys in $^4C_3$ ways.

Therefore, by the multiplication principle, the number of ways in which a team of 3 boys and 3 girls can be selected $=^5C_3\times ^4C_3$

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

$=10\times 4=40$

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 $^6C_3$.

3 balls are selected from 5 white balls in $^5C_3$

3 balls are selected from 5 blue balls in $^5C_3$.

Hence, by the multiplication principle, the number of ways of selecting 9 balls $=^6C_3\times ^5C_3\times ^5C_3$

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

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

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 $^4C_1$ 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 $^5C_2$ ways and 3 red balls can be selected in $^6C_3$ ways.

Thus, using multiplication priciple, number of ways of selecting 2 black and 3 red balls $=^5C_2 .^6C_3$

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

$=10\times 20=200$

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 $^7C_3$ ways.

Thus, using the multiplication principle, the number of ways of selecting courses $=^7C_3$

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

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

$=35$