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

The correct option is D $3{{}^{m+1}C}_4$

$\alpha ={}^m{C}_2$

${}^{\alpha }{C}_2=?$ (say $\beta$)

$\alpha =\frac{m!}{(m-2)!2!}=\frac{m(m-1)}{2!}$

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

$=\frac{\left(\frac{m(m-1)}{2}\right)\left(\frac{m(m-1)-2}{2}\right)}{2}$

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

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

$\frac{1}{8}\frac{(m+1)!}{(m-3)!}=\frac{3}{\mathrm{1.2.3.4}}\frac{(m+1)!}{(m-3)!}$

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