Question

${}^8{P}_5+{5.}^8{P}_4{=}^{9}P{}_r$

Answer 1

We have,

We know that,

${}^n{P}_r=\frac{n!}{(n-r)!}$

$now,$

${}^8{P}_5+{5.}^8{P}_4{=}^9{P}_r$

$\Rightarrow \frac{8!}{(8-5)!}+5\times \frac{8!}{(8-4)!}=\frac{9!}{(9-r)!}$

$\Rightarrow \frac{8!}{3!}+5\times \frac{8!}{4!}=\frac{9\times 8!}{(9-r)!}$

$\Rightarrow 8!(\frac{1}{6}+\frac{5}{24})=\frac{9\times 8!}{(9-r)!}$

$\Rightarrow \frac{4+5}{24}=\frac{9}{(9-r)!}$

$\Rightarrow \frac{9}{24}=\frac{9}{(9-r)!}$

$\Rightarrow \frac{1}{24}=\frac{1}{(9-r)!}$

$\Rightarrow (9-r)!=24$

$\Rightarrow (9-r)!=3!$

$\Rightarrow 9-r=3$

$\Rightarrow -r=3-9$

$\Rightarrow -r=-6$

$\Rightarrow r=6$

Hence, this is the answer.