MCQ
$^n{P_r}$ =
- ✓$^{n - 1}{P_r} + r{\,^{n - 1}}{P_{r - 1}}$
- B$n.{\;^{n - 1}}{P_r}{ + ^{n - 1}}{P_{r - 1}}$
- C$n{(^{n - 1}}{P_r}{ + ^{n - 1}}{P_{r - 1}})$
- D$^{n - 1}{P_{r - 1}}{ + ^{n - 1}}{P_r}$
$ = \frac{{(n - 1)\,!}}{{(n - 1 - r)\,!}} + r\frac{{(n - 1)\,!}}{{(n - r)\,!}}$
$\left( {\because \,\,{\,^n}{P_r} = \frac{{n\,!}}{{(n - r)\,!}}} \right)$
= $\frac{{(n - 1)\,!}}{{(n - 1 - r)\,!}}\,\,\left\{ {1 + r.\frac{1}{{n - r}}} \right\}$
= $\frac{{(n - 1)\,!}}{{(n - 1 - r)\,!(n - r)\,!}}\left( {\frac{n}{{n - r}}} \right) = \frac{{n\,!}}{{(n - r)\,!}} = {\,^n}{P_r}$.
Aliter : We know that $^{n - 1}{C_r} + {\,^{n - 1}}{C_{r - 1}} = {\,^n}{C_r}$
==> $\frac{{^{n - 1}{P_r}}}{{r\,!}} + \frac{{^{n - 1}{P_{r - 1}}}}{{(r - 1)\,!}} = \frac{{^n{P_r}}}{{r\,!}}$
==> $^{n - 1}{P_r} + r\,.{\,^{n - 1}}{P_{r - 1}} = {\,^n}{P_r}$.
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