Mathematics - Binomial Theorem Question with Solution | TestHub

$n={}^m{\mathrm{C}}_2=\frac{m(m-1)}{2}$ Since $m$ and $(m-1)$ are two consecutive natural numbers, therefore their product is an even natural number. So $\frac{m(m-1)}{2}$ is also a natural number. Now $\frac{m(m-1)}{2}=\frac{m^2-m}{2}$ $$\begin{array}{rl}\therefore & \frac{m(m-1)}{2}{\mathrm{C}}_2=\frac{\left(\frac{m^2-m}{2}\right)\left(\frac{m^2-m}{2}-1\right)}{2}\\ & =\frac{m(m-1)\left(m^2-m-2\right)}{8}\\ & =\frac{m(m-1)\left[m^2-2m+m-2\right]}{8}\\ & =\frac{m(m-1)[m(m-2)+1(m-2)]}{8}\\ & =\frac{m(m-1)(m-2)(m+1)}{8}\end{array}$$