If {}^n P_r = {}^{(n-1)} P_r + x cdot {}^{(n- | vedclass

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(D) We know that ${}^n P_r = \frac{n!}{(n-r)!}$.Consider the expression ${}^n P_r - {}^{n-1} P_r = \frac{n!}{(n-r)!} - \frac{(n-1)!}{(n-r-1)!}$.$= \fr

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$r$

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(D) We know that ${}^n P_r = \frac{n!}{(n-r)!}$.Consider the expression ${}^n P_r - {}^{n-1} P_r = \frac{n!}{(n-r)!} - \frac{(n-1)!}{(n-r-1)!}$.$= \frac{(n-1)!}{(n-r-1)!} \left( \frac{n}{n-r} - 1 \right) = \frac{(n-1)!}{(n-r-1)!} \left( \frac{n - (n-r)}{n-r} \right)$.$= \frac{(n-1)!}{(n-r-1)!} \cdot \frac{r}{n-r} = \frac{r \cdot (n-1)!}{(n-r)!}$.$= r \cdot \frac{(n-1)!}{((n-1) - (r-1))!} = r \cdot {}^{n-1} P_{r-1}$.Thus,${}^n P_r = {}^{n-1} P_r + r \cdot {}^{n-1} P_{r-1}$.Comparing this with the given equation ${}^n P_r = {}^{n-1} P_r + x \cdot {}^{n-1} P_{r-1}$,we get $x = r$.

If ${}^n P_r = {}^{(n-1)} P_r + x \cdot {}^{(n-1)} P_{(r-1)}$,$\forall n, r \in N$ and $r \leq n$,then $x =$

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