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

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(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

Vedclass · Accepted Answer

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

Vedclass · Answer

(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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