If ${ }^9{P}_5+5\cdot {}^9{P}_4={}^{10}{P}_r$, then $\text{'}r\text{'}=$

L.H.S. $={}^9{P}_5+5\cdot {}^9{P}_4==\frac{9!}{(9-5)!}+5\frac{9!}{(9-4)!}=\frac{9!}{4!}+5\frac{9!}{5!}$

$=\frac{9!}{4!}+\frac{5\cdot 9!}{5\times 4!}=\frac{9!}{4!}+\frac{9!}{4!}$

$=2\cdot \frac{9!}{4!}=\frac{5\times 2\times 91}{5\times 4!}=\frac{10\times 9!}{5\times 4!}$

$\therefore \text{L.H.S.}=\frac{10!}{5!}$

$\therefore \text{R.H.S.}=\frac{10!}{(10-r)!}$

$L.H.S.=R.H.S.$ (given)

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

$\Rightarrow 10-r=5\Rightarrow r=5$.