Question

If $\alpha {=}^m{C}_2$, then ${}^{\alpha }{C}_2$ is equal to

• A. ${}^{m+1}{C}_4$
• B. ${}^{m-1}{C}_4$
• C. $3{}^{m+2}{C}_4$
• D. $3{}^{m+1}{C}_4$

Answer 1

Answer: (D)

$3{}^{m+1}{C}_4$

$\alpha {=}^m{C}_2\Rightarrow \alpha =\frac{m(m-1)}{2}$

${\therefore }^a{C}_2=\frac{\alpha (\alpha -1)}{2}=\frac{1}{2}\frac{m(m-1)}{2}\{\frac{m(m-1)}{2}-1\}$

$=\frac{1}{8}m(m-1)(m-2)(m+1)$

$=\frac{1}{8}(m+1)m(m-1)(m-2)=3{}^{m+1}{C}_4$.