Question

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

• A. $1$
• B. $2$
• C. $2n-1$
• D. $2n-2$

Answer 1

Answer: (C)

$2n-1$

$\frac{n!}{(n-r)!}=\frac{n!}{(n-s)!}$

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

Given $r<s⟹-r>-s$

$\therefore (n-r)>(n-s)$

We know that two different factorials are zero and one

$\therefore n-r=1andn-s=0$

$\therefore r=n-1,s=n$

$\therefore r+s=2n-1$