If $n={ }^m C_2$, then the value of ${ }^n C_2$ is given by

• Correct answer
  A

  $3\left({ }^{m+1} C_4\right)$
• B

  ${ }^{m-1} C_4$
• C

  ${ }^{m+1} C_4$
• D

  $2\left({ }^{m+2} C_4\right)$

Solution

$n={ }^m \mathrm{C}_2=\frac{m(m-1)}{2}$ 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 $\frac{m(m-1)}{2}=\frac{m^2-m}{2}$ $$ \begin{aligned} \therefore & \frac{m(m-1)}{2} \mathrm{C}_2=\frac{\left(\frac{m^2-m}{2}\right)\left(\frac{m^2-m}{2}-1\right)}{2} \\ & =\frac{m(m-1)\left(m^2-m-2\right)}{8} \\ & =\frac{m(m-1)\left[m^2-2 m+m-2\right]}{8} \\ & =\frac{m(m-1)[m(m-2)+1(m-2)]}{8} \\ & =\frac{m(m-1)(m-2)(m+1)}{8} \end{aligned} $$

Hence, the correct answer is option: 'A'.