6. permutation and combination — ગણિત ધોરણ 11 સાયન્સ — Question

$ = \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 $^{n - 1}{C_r} + {\,^{n - 1}}{C_{r - 1}} = {\,^n}{C_r}$

==> $\frac{{^{n - 1}{P_r}}}{{r\,!}} + \frac{{^{n - 1}{P_{r - 1}}}}{{(r - 1)\,!}} = \frac{{^n{P_r}}}{{r\,!}}$

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