Question

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

• A. $m+1C_4$
• B. $m-1C_4$
• C. $m+2C_4$
• D. None of these

Answer 1

The correct option is D None of these

$\begin{array}{c}given:-\\ n={}^m{C}_2=\\ \frac{m!}{2!\times \left(m-2\right)!}=\frac{m\left(m-1\right)\left(m-2\right)!}{2\times \left(m-2\right)!}=\frac{m\left(m-1\right)}{2}\\ Thus,\\ {}^n{C}_2=\frac{a!}{2!\times \left(a-2\right)!}=\frac{a\left(a-1\right)\left(a-2\right)!}{2\times \left(a-2\right)!}=\frac{a\left(a-1\right)}{2}\\ =\frac{\left(\frac{m\left(m-1\right)}{2}\right)\left(\frac{m\left(m-1\right)}{2}-1\right)}{2}\\ =\frac{m\left(m-1\right)\left(m^2-m-2\right)}{8}\end{array}$

Hence,

option $D$ Iis correct answer.