Question

If $0 < r < s \leqslant n$and ${}^n{P_r} = {}^n{P_s}$, then the value of $r + s$ is:
$\left( A \right).$ 1
$\left( B \right).$ 2
$\left( C \right).$ $2n - 1$
$\left( D \right).$ $2n - 2$

Answer 1

Hint: Use formulas of permutation to find the value.

We know that:
${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}{\text{ }}\left( {{\text{Permutation Formula}}} \right)$
Given that: ${}^n{P_r} = {}^n{P_s}$
$
  \therefore \dfrac{{n!}}{{\left( {n - r} \right)!}} = \dfrac{{n!}}{{\left( {n - s} \right)!}} \\
  \left( {n - r} \right)! = \left( {n - s} \right)! \\
$
Also, $r < s{\text{ }}\left( {{\text{Given}}} \right)$
$\therefore - r > - s$
Adding $n$both sides, we get
$\left( {n - r} \right) > \left( {n - s} \right)$
We know that two different factorials having the same value are 0 and 1, both having factorial equal to 1.
$\therefore n - r = 1$and $n - s = 0$
$
   \Rightarrow r = n - 1,s = n \\
  \therefore r + s = n + n - 1 \\
  r + s = 2n - 1 \\
$
Hence, the correct option is C.

Note: Whenever you see permutation, always try to expand the term by using a permutation formula which makes calculation easy.