Exercise 6.4 (Revised) - Chapter 7 - Permutations & Combinations - Ncert Solutions class 11 - Maths
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Chapter 6: Permutations & Combinations - NCERT Solutions Class 11 Maths
Ex 6.4 Question 1.
If ${ }^n \mathrm{C}_8={ }^n \mathrm{C}_2$, find ${ }^n \mathrm{C}_2$
Answer.
Given: ${ }^n \mathrm{C}_8={ }^n \mathrm{C}_2$
$
\begin{aligned}
& \Rightarrow{ }^n \mathrm{C}_{\mathbb{8}}={ }^n \mathrm{C}_{n-2} \\
& \Rightarrow 8=n-2 \\
& \Rightarrow n=10 \\
& \therefore{ }^n \mathrm{C}_2={ }^{10} \mathrm{C}_2=\frac{10!}{2!8!}=45
\end{aligned}
$
Ex 6.4 Question 2.
Determine if:
(i) ${ }^{2 n} \mathrm{C}_3 \cdot n \mathrm{C}_2=12: 1$
(ii) ${ }^{2 n} \mathrm{C}_3 \cdot{ }^n \mathrm{C}_3=11: 1$
Answer.
(i) Given: ${ }^{2 n} \mathrm{C}_3:{ }^n \mathrm{C}_2=12: 1$
$
\begin{aligned}
& \Rightarrow \frac{(2 n)!}{3!(2 n-3)!} \times \frac{2!(n-2)!}{n!}=\frac{12}{1} \\
& \Rightarrow \frac{(2 n)(2 n-1)(2 n-2)(2 n-3)!}{3 \times 2!(2 n-3)!} \times \frac{2!(n-2)!}{n(n-1)(n-2)!}=\frac{12}{1}
\end{aligned}
$
$\Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{3} \times \frac{1}{n(n-1)}=\frac{12}{1}$
$\begin{aligned}
&\begin{aligned}
& \Rightarrow \frac{4(2 n-1)}{3}=\frac{12}{1} \\
& \Rightarrow 8 n-4=36 \\
& \Rightarrow n=5
\end{aligned}\\
&\begin{aligned}
& \text { (ii) Given: }{ }^{2 n} \mathrm{C}_3{ }^n \mathrm{C}_3=11: 1 \\
& \Rightarrow \frac{(2 n)!}{3!(2 n-3)!} \times \frac{2!(n-3)!}{n!}=\frac{11}{1} \\
& \Rightarrow \frac{(2 n)(2 n-1)(2 n-2)(2 n-3)!}{3!(2 n-3)!} \times \frac{3!(n-3)!}{n(n-1)(n-2)(n-3)!}=\frac{11}{1} \\
& \Rightarrow \frac{(2 n)(2 n-1)(2 n-2)}{n(n-1)(n-2)}=\frac{11}{1} \\
& \Rightarrow \frac{4(2 n-1)}{(n-2)}=\frac{11}{1} \\
& \Rightarrow 8 n-4=11 n-22 \\
& \Rightarrow 3 n=18 \\
& \Rightarrow n=6
\end{aligned}
\end{aligned}$
Ex 6.4 Question 3.
How many chords can be drawn through 21 points on a circle?
Answer.
There are 21 points on the circumference of a circle. Since one and only one chord can be drawn by joining 2 distinct points, therefore the required number of chords given by
$
{ }^{21} \mathrm{C}_2=\frac{21!}{2!19!}=\frac{21 \times 20 \times 19!}{2!19!}=210
$
Ex 6.4 Question 4.
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Answer.
There are 5 boys and 4 girls and 3 out of 5 boys and 3 out of 4 girls have to be selected.
$
\begin{aligned}
& \therefore \text { Number of ways of selection }={ }^5 \mathrm{C}_3 \times{ }^4 \mathrm{C}_3 \\
& =\frac{5!}{3!2!} \times \frac{4!}{3!1!}=10 \times 4=40
\end{aligned}
$
Ex 6.4 Question 5.
Find the number 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.
Answer.
There are 6 red balls, 5 white balls and 5 blue balls. 3 balls of each colour have to be found.
$
\begin{aligned}
& \therefore \text { Number of ways of selection }={ }^6 \mathrm{C}_3 \times{ }^5 \mathrm{C}_3 \times{ }^5 \mathrm{C}_3 \\
& =\frac{6!}{3!3!} \times \frac{5!}{3!2!} \times \frac{5!}{3!2!} \\
& =20 \times 10 \times 10=2000
\end{aligned}
$
Ex 6.4 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.
There are 4 aces and 48 other cards in a deck of 52 cards. We have to select 1 ace out of 4 aces and 4 other cards out of 48 other cards.
$
\begin{aligned}
& \therefore \text { Number of ways of selection }={ }^4 \mathrm{C}_1 \times{ }^{48} \mathrm{C}_4 \\
& =\frac{4!}{1!3!} \times \frac{48!}{44!4!} \\
& =\frac{4 \times 3!}{3!} \times \frac{48 \times 47 \times 46 \times 45 \times 44!}{4 \times 3 \times 2 \times 44!}=778320
\end{aligned}
$
Ex 6.4 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.
There are 5 bowlers and 12 other players in a team of 17 players we have to select 4 bowels out of 5 bowlers and 7 other players out of 12 other players.
$
\begin{aligned}
& \therefore \text { Number of ways of selection }={ }^5 \mathrm{C}_4 \times{ }^{12} \mathrm{C}_7 \\
& =\frac{5!}{1!4!} \times \frac{12!}{7!5!} \\
& =\frac{5 \times 4!}{4!} \times \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{5 \times 4 \times 3 \times 2 \times 1 \times 7!} \\
& =5 \times 792=3960
\end{aligned}
$
Ex 6.4 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.
There are 5 black and 6 red balls. We have to select 2 black balls out of 5 black balls and 3 red balls out of 6 red balls.
$\therefore$ Number of ways of selection $={ }^5 \mathrm{C}_2 \times{ }^6 \mathrm{C}_3=\frac{5!}{2!3!} \times \frac{6!}{3!3!}$
$
=\frac{5 \times 4 \times 3!}{2!3!} \times \frac{6 \times 5 \times 4 \times 3!}{3!3!}=200
$
Ex 6.4 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.
There are 9 courses and number of courses to be selected are 5 in which 2 specific courses are compulsory. 3 courses out of remaining 7 courses have to be selected.
$
\begin{aligned}
& \therefore \text { Number of ways of selection }={ }^7 \mathrm{C}_3=\frac{7!}{4!3!} \\
& =\frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!}=35
\end{aligned}
$
