Question

If $m{=}^n{\text{C}}_2$, then ${}^m{\text{C}}_2$ equals

• A. ${}^{n+1}{\text{C}}_4$
• B. $3{\times }^{n+1}{\text{C}}_4$
• C. ${}^n{\text{C}}_4$
• D. ${}^{n+1}{\text{C}}_3$

Answer 1

The correct option is B $3{\times }^{n+1}{\text{C}}_4$

$m{=}^n{C}_2=\frac{n(n-1)}{2}...(\ast )$
${\therefore }^m{C}_2=\frac{m(m-1)}{2}=\frac{\frac{n(n-1)}{2}[\frac{n(n-1)}{2}-1]}{2}$ using (*)
$=\frac{n(n-1)(n^2-n-2)}{8}=\frac{n(n-1)(n-2)(n+1)}{8}$

$=\frac{3(n+1)\left(n\right)(n-1)(n-2)}{4!}$

$={3.}^{n+1}{C}_4$