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STD 11 - 6. permutation and combination — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 6. permutation and combination4 Marks
MCQ
The value of $^n{P_r}$ is equal to
  • ✓
    $^{n - 1}{P_r} + r{\,^{n - 1}}{P_{r - 1}}$
  • B
    $n.{\;^{n - 1}}{P_r}{ + ^{n - 1}}{P_{r - 1}}$
  • C
    $n{(^{n - 1}}{P_r}{ + ^{n - 1}}{P_{r - 1}})$
  • D
    $^{n - 1}{P_{r - 1}}{ + ^{n - 1}}{P_r}$
✓

Answer

Correct option: A.
$^{n - 1}{P_r} + r{\,^{n - 1}}{P_{r - 1}}$
a
(a) $^{n - 1}{P_r} + r{.^{n - 1}}{P_{r - 1}}$

$ = \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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