Daily Question 107

Daily Question 107

$\textbf{Q107.}$ If $n={^{m}}C_{2},$ then the value of ${^{n}}C_{2}$ is

(A) $3\cdot{^{m+1}}C_{4}$

(B) ${^{m+2}}C_{4}$

(C) ${^{m+3}}C_{4}$

(D) $m\cdot{^{m}}C_{4}$

$\textbf{Ans.} (A)$

$\textbf{Sol.}$ Given $n={^{m}}C_{2}$

$\implies n=\dfrac{m(m-1)}{2}$

Now, ${^{n}}C_2=\dfrac{n(n-1)}{2}=\dfrac{1}{2}\dfrac{m(m-1)}{2}\left[\dfrac{m(m-1)}{2}-1\right]$

$\implies {^{n}}C_2 = \dfrac{1}{8}m(m-1)(m^2-m-2)$

$\implies {^{n}}C_2 = \dfrac{3}{4!}(m+1)m(m-1)(m-2)$

$\implies {^{n}}C_2 = 3\cdot{^{m+1}}C_4$

Compiled Daily Questions 1-100