(B) Given the condition $0 < r < s < n$ and ${}^n P_r = {}^n P_s$. Since ${}^n P_r = \frac{n!}{(n - r)!}$ and ${}^n P_s = \frac{n!}{(n - s)!}$,we have

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(B) Given the condition $0 Since ${}^n P_r = \frac{n!}{(n - r)!}$ and ${}^n P_s = \frac{n!}{(n - s)!}$,we have $\frac{n!}{(n - r)!} = \frac{n!}{(n - s)!}$. This implies $(n - r)! = (n - s)!$. Since $r (n - s)$. For the factorials of two distinct non-negative integers to be equal,the only possibility is $0! = 1! = 1$. Thus,we must have $n - s = 0$ and $n - r = 1$. From $n - s = 0$,we get $s = n$. From $n - r = 1$,we get $r = n - 1$. Therefore,$r + s = (n - 1) + n = 2n - 1$.

Class 11 Mathematics Permutation and Combination Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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If $0 < r < s < n$ and ${}^n P_r = {}^n P_s$,then find the value of $r + s$.

TS EAMCET 2021 All TS EAMCET

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

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Class 11

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Chapter

Permutation and Combination

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Definition of permutation, Number of permutations with or without repetition, Conditional permutations

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If $0 < r < s < n$ and ${}^n P_r = {}^n P_s$,then find the value of $r + s$.

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