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Question: If \(0 < r < s \leqslant n\)and \({}^n{P_r} = {}^n{P_s}\), then the value of \(r + s\) is: \(\left...

If 0<r<s⩽n0 < r < s \leqslant nand nPr=nPs{}^n{P_r} = {}^n{P_s}, then the value of r+sr + s is:
(A).\left( A \right). 1
(B).\left( B \right). 2
(C).\left( C \right). 2n−12n - 1
(D).\left( D \right). 2n−22n - 2

Explanation

Solution

Hint: Use formulas of permutation to find the value.

We know that:
nPr=n!(n−r)! (Permutation Formula){}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}{\text{ }}\left( {{\text{Permutation Formula}}} \right)
Given that: nPr=nPs{}^n{P_r} = {}^n{P_s}
∴n!(n−r)!=n!(n−s)! (n−r)!=(n−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 (Given)r < s{\text{ }}\left( {{\text{Given}}} \right)
∴−r>−s\therefore - r > - s
Adding nnboth sides, we get
(n−r)>(n−s)\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.
∴n−r=1\therefore n - r = 1and n−s=0n - s = 0
⇒r=n−1,s=n ∴r+s=n+n−1 r+s=2n−1  \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.