If $n{=}^m{C}_2$, the value of ${ }^n{C}_2$ is given  by

We have,

$\begin{array}{c}n{=}^m{C}_2=\frac{m+n-1}{2}\\ \therefore { }^n{C}_2=\frac{n\left(n-1_1\right.}{2}\end{array}$

$\begin{array}{l}{\Rightarrow }^n{C}_2=\frac{m(m-1)}{4}\left\{\frac{m(m-1)}{2}-1\right\}\\ \Rightarrow {}^n{C}_2=\frac{1}{8}m(m-1)\left(m^2-m-2\right)\\ \Rightarrow {}^n{C}_2=\frac{1}{8}m(m-1)(m-2)(m+1)\\ \Rightarrow {}^n{C}_2=3\left\{\frac{1}{24}(m+1)m(m-1)(m-2)\right\}=3{\times }^{m+3}{C}_4\end{array}$