Question

lf $m{=}^n{C}_2$, then ${}^m{C}_2$ equals

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

Answer 1

Answer: (B)

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

$m{=}^n{C}_2=\frac{n(n-1)}{2}=\frac{n^2-n}{2}$
${\therefore }^m{C}_2=\frac{m(m-1)}{2}=\frac{1}{2}\frac{n^2-n}{2}(\frac{n^2-n}{2}-1)$
$=\frac{n(n-1)(n^2-n-2)}{8}=\frac{(n+1)\left(n\right)(n-1)(n-2)}{8}$
$=3\cdot \frac{(n+1)!}{4!(n-3)!}=3{\cdot }^{n+1}{C}_4$