Question

If $8! \begin{bmatrix}\frac{1}{3!}+\frac{5}{4!}\end{bmatrix} =^{9}P_{r} ,$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$