Q. The value of $\ ^{n}P_{r}$is equal to
A) $\ ^{n - 1}P_{r} + r.\ ^{n - 1}P_{r - 1}$
B) ${n.\ }^{n - 1}P_{r} + \ ^{n - 1}P_{r - 1}$
C) ${n(\ }^{n - 1}P_{r} + \ ^{n - 1}P_{r - 1})$
D) $\ ^{n - 1}P_{r - 1} + \ ^{n - 1}P_{r}$
Explanation
(a)
Taking option (a)
$\ ^{n - 1}P_{r} + r\ ^{n - 1}P_{r - 1} = \frac{(n - 1)!}{(n - 1 - r)!} + \frac{(n - 1)!}{(n - r)!}$
$\left( \therefore{\ ^{n}P}_{r} = \frac{n!}{(n - r)!} \right)$
$= \frac{(n - 1)!}{(n - 1 - r)!}\left( 1 + r.\frac{1}{n - r} \right)$
$= \frac{(n - 1)!}{(n - 1 - r)!}\left( \frac{n}{n - r} \right) = \frac{n!}{(n - r)!} = {\ ^{n}P}_{r}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.