If $0<r<s\le n$ and ${ }^n{P}_r{=}^n{P}_s$ then value  of $r + s$is

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

$\Rightarrow (n-r)!=(n-s)!$

As $r<s,n-r>n-s$ But the only two different factorials which are equal are $0!$ and $1!$ Thus $n – r = 1$ and $n – s = 0$ 

$\begin{array}{r}\Rightarrow r=n-1\text{ and }s=n.\\ \Rightarrow r+s=2n-1.\end{array}$