Question

If ${}^9{P}_5+5{}^9{P}_4={}^{10}{P}_r,$ then value of $r$ is

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is C $5$

${}^{10}{P}_r={}^9{P}_5+5{}^9{P}_4$
$({}^n{P}_r=\frac{n!}{(n-r)!})$

$\therefore {}^{10}{P}_r=\frac{9!}{(9-5)!}+5\times \frac{9!}{(9-4)!}=\frac{9!}{4!}+5\times \frac{9!}{5!}=\frac{9!}{4!}+\frac{9!}{4!}=2\times \frac{9!}{4!}=\frac{5\times 2\times 9!}{5\times 4!}=\frac{10\times 9!}{5!}=\frac{10!}{5!}={}^{10}{P}_5\Rightarrow {}^{10}{P}_r={}^{10}{P}_5\Rightarrow r=5$

Alternate :
${}^{10}{P}_r={}^9{P}_5+5{}^9{P}_4$
$({}^n{P}_r+r{}^n{P}_{r-1}={}^{n+1}{P}_r)$
$\Rightarrow {}^{10}{P}_r={}^{10}{P}_5$
$\therefore r=5$