Q. If $m = \ ^{n}C_{2}$, then $\ ^{m}C_{2}$ is equal to
A) $3\ ^{n}C_{4}\$
B) $\ ^{n + 1}C_{4}$
C) $3^{.n + 1}C_{4}$
D) $3.\ ^{n + 1}C_{3}$
Explanation
(c)
Given, $m = {\ ^{n}C}_{2} = \frac{n!}{2!(n - 2)!} = \frac{n(n - 1)}{2}$
Now, ${\ ^{n}C}_{2} = \frac{m!}{2!(m - 2)!} = \frac{m(m - 1)}{2}$
$= \frac{\frac{n(n - 1)}{2}.\left( \frac{n^{2} - n - 2}{2} \right)}{2}$
$= \frac{(n + 1)n(n - 1)(n - 2)}{8}$
$= 3.{\ ^{n + 1}C}_{4}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.