Question

If $8!\left[\begin{array}{c}\frac{1}{3!}+\frac{5}{4!}\end{array}\right]{=}^9{P}_r,$then the value of $r$ is equal to

• A. 4
• B. 5
• C. 3
• D. 2
• E. 1

Answer 1

Answer: (B)

We have, $8!\left[\frac{1}{3!}+\frac{5}{4!}\right]={}^9{P}_r$
$\Rightarrow 8!\left[\frac{1}{3!}+\frac{b}{4\cdot (3!)}\right]={}^9{P}_r$
$\Rightarrow \frac{8!}{3!}\left(1+\frac{5}{4}\right)={}^9{P}_r$
$\Rightarrow \frac{8!9}{3!4}={}^9{P}_r$
$\Rightarrow \frac{9!}{4!}=\frac{9!}{(9-r)!}$
On comparing, we get
$9-r=4$
$\Rightarrow r=5$