# 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

$\left( C \right).$ $2n - 1$

$\left( D \right).$ $2n - 2$

Last updated date: 16th Mar 2023

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Answer

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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.

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.

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