If  $o<r<s\le n$and  $n_{P_r}=n_{P_s}$ then $r+s$ =

$\begin{array}{rl} & 0<\mathrm{r}<\mathrm{s}<\mathrm{n}\Rightarrow \\ & \Rightarrow \frac{n!}{(n-r)!}=\frac{n!}{(n-s)!}\\ & \Rightarrow (n-r)!=(n-s)!\\ & \text{As,}0<r<s<n\Rightarrow (n-r)>(n-s)\end{array}$
So, the only possibility to hold this equality is n – r= 1 and n – s = 0
$\Rightarrow r=n-1\text{and }s=n\Rightarrow r+s=2n-1\text{.}$