Exercise 6.2 (Revised) - Chapter 7 - Permutations & Combinations - Ncert Solutions class 11 - Maths

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Chapter 6: Permutations & Combinations - NCERT Solutions Class 11 Math

Ex 6.2 Question 1.

Evaluate:
(i) 8 !
(ii) (ii) 4 ! -3 !

Answer.

(i) 8 ! $=8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=40320$
(ii) 4 ! -3 ! $=4 \times 3 \times 2 \times 1-3 \times 2 \times 1=24-6=18$
Ex 6.2 Question 2.

Is $3!+4!=7!?$

Answer.
$\mathrm{LHS}=3!+4!=3 \times 2 \times 1+4 \times 3 \times 2 \times 1$
$
=6+24=30
$
$\mathrm{RHS}=7!=7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=5040$
$\therefore \mathrm{LHS} \neq \mathrm{RHS}$
No,
Ex 6.2 Question 3.

Compute: $\frac{8!}{6!\times 2!}$

Answer.

Given : $\frac{8!}{6!\times 2!}=\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 2 \times 1}=\frac{8 \times 7}{2}=28$.

Ex 6.2 Question 4.

If $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$ find $x$.

Answer.

Given: $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$
$
\Rightarrow \frac{1}{6!}+\frac{1}{7 \times 6!}=\frac{x}{8 \times 7 \times 6!}
$

$
\begin{aligned}
& \Rightarrow \frac{1}{6!}\left[1+\frac{1}{7}\right]=\frac{x}{8 \times 7 \times 6!} \\
& \Rightarrow \frac{8}{7}=\frac{x}{8 \times 7} \Rightarrow x=8 \times 8=64
\end{aligned}
$
Ex 6.2 Question 5.

Evaluate: $\frac{n!}{(n-r)!}$ when
(i) $n=6, r=2$
(ii) $n=9, r=5$

Answer.

(i) Given: $n=6$ and $r=2$
$
\begin{aligned}
& \therefore \frac{n!}{(n-r)!}=\frac{6!}{(6-2)!}=\frac{6!}{4!} \\
& =\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}=6 \times 5=30 .
\end{aligned}
$
(ii) Given: $n=9$ and $r=5$
$
\begin{aligned}
& \therefore \frac{n!}{(n-r)!}=\frac{9!}{(9-5)!}=\frac{9!}{4!} \\
& =\frac{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{4 \times 3 \times 2 \times 1}=9 \times 8 \times 7 \times 6 \times 5=15120 .
\end{aligned}
$