Question

If ${}^9{P}_5$ + 5. ${}^9{P}_4$ = ${}^1{}^0{P}_r$, then r$=$

• A. $4$
• B. $5$
• C. $6$
• D. $7$

Answer 1

The correct option is B $5$

$LHS{\to }^n{P}_r+r^n{P}_{r-1}$
$\Rightarrow \frac{n!}{(n-r)!}+r\frac{n!}{(n-r+1)!}$
$\Rightarrow \frac{(n-r+1)n!}{(n-r+1)!}+\left(r\right)\frac{n!}{(n-r+1)!}$
$\Rightarrow \frac{n!}{(n-r+1)!}(n-r+1+r)$
$\Rightarrow \frac{(n+1)n!}{(n-r+1)!}=\frac{(n+1)!}{(n-r+1)!}{=}^{n+1}{P}_r$
Here, $n=9$
$r=5$
So, RHS $=$ ${}^{n+1}{P}_r{=}^{10}{P}_5$