The value of ^n{P_r} is equal to | vedclass

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Question

(A) The correct option is $A$.We know that $^n{P_r} = \frac{n!}{(n-r)!}$.Consider the expression in option $A$: $^{n-1}{P_r} + r \cdot {^{n-1}}{P_{r-1

Vedclass · Accepted Answer

$^{n - 1}{P_r} + r{\,^{n - 1}}{P_{r - 1}}$

Vedclass · Answer

(A) The correct option is $A$.We know that $^n{P_r} = \frac{n!}{(n-r)!}$.Consider the expression in option $A$: $^{n-1}{P_r} + r \cdot {^{n-1}}{P_{r-1}}$.$= \frac{(n-1)!}{(n-1-r)!} + r \cdot \frac{(n-1)!}{(n-1-(r-1))!}$$= \frac{(n-1)!}{(n-r-1)!} + r \cdot \frac{(n-1)!}{(n-r)!}$$= \frac{(n-1)!}{(n-r-1)!} \left[ 1 + \frac{r}{n-r} \right]$$= \frac{(n-1)!}{(n-r-1)!} \left[ \frac{n-r+r}{n-r} \right]$$= \frac{(n-1)!}{(n-r-1)!} \cdot \frac{n}{n-r}$$= \frac{n \cdot (n-1)!}{(n-r) \cdot (n-r-1)!} = \frac{n!}{(n-r)!} = ^n{P_r}$.

The value of $^n{P_r}$ is equal to

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