(n - r + 1) times ^nP_{r - 1} = dots | vedclass

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Question

(C) We know that the formula for permutation is $^nP_r = \frac{n!}{(n-r)!}$.Consider the expression $(n - r + 1) 	imes ^nP_{r - 1}$.Using the formula

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$^nP_r$

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(C) We know that the formula for permutation is $^nP_r = \frac{n!}{(n-r)!}$.Consider the expression $(n - r + 1) 	imes ^nP_{r - 1}$.Using the formula,$^nP_{r-1} = \frac{n!}{(n - (r - 1))!} = \frac{n!}{(n - r + 1)!}$.Substituting this into the expression:$(n - r + 1) 	imes \frac{n!}{(n - r + 1)!} = (n - r + 1) 	imes \frac{n!}{(n - r + 1) 	imes (n - r)!}$.Canceling the term $(n - r + 1)$ from the numerator and denominator:$= \frac{n!}{(n - r)!} = ^nP_r$.Thus,the correct option is $C$.

$(n - r + 1) \times ^nP_{r - 1} = \dots$

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