Question

If ${}^9{P}_5+{5.}^9{P}_4{=}^{10}{P}_r$ find $r$.

Answer 1

${9_P}_5+5{9_P}_4={{10}_P}_r$

$\Rightarrow \frac{9!}{(9-5)!}+5\frac{9!}{(9-4)!}=\frac{10!}{(10-r)!}$ using the formula ${n_P}_r=\frac{n!}{(n-r)!}$

$\Rightarrow \frac{9!}{4!}+5\frac{9!}{5!}=\frac{10\times 9!}{(10-r)!}$

$\Rightarrow \frac{1}{4!}+5\frac{1}{5!}=\frac{10}{(10-r)!}$ by dividing both sides by $9!$

$\Rightarrow \frac{5}{5!}+\frac{5}{5!}=\frac{10}{(10-r)!}$

$\Rightarrow \frac{10}{5!}=\frac{10}{(10-r)!}$

$\Rightarrow 5=10-r$

$\Rightarrow -r=5-10$

$\therefore r=5$