Question

If $m={}^{n}C_2$, then ${}^{m}C_2$ is equal to

• A. $3·{}^{n}C_4$
• B. ${}^{n+1}C_4$
• C. $3·{}^{n+1}C_4$
• D. $3·{}^{n+1}C_3$
• E. $3·{}^{n+1}C_2$

Answer 1

The correct option is C

$3·{}^{n+1}C_4$

The explanation for the correct option

It is given that $m={}^{n}C_2$.

$$\begin{gathered} \Rightarrow m=\frac{n!}{2!\times \left(n-2\right)!} \\ \Rightarrow m=\frac{n\left(n-1\right)}{2} \end{gathered}$$

Thus, ${}^{m}C_2=\frac{m!}{2!\times \left(m-2\right)!}$

$$\begin{gathered} \Rightarrow {}^{m}C_2=\frac{m\left(m-1\right)}{2} \\ \Rightarrow {}^{m}C_2=\frac{{\displaystyle \frac{n\left(n-1\right)}{2}}\left[{\displaystyle \frac{n\left(n-1\right)}{2}}-1\right]}{2} \\ \Rightarrow {}^{m}C_2=\frac{{\displaystyle \frac{n\left(n-1\right)}{2}\left[\frac{n^2-n-2}{2}\right]}}{2} \\ \Rightarrow {}^{m}C_2=\frac{{\displaystyle \frac{n\left(n-1\right)}{2}\left[\frac{n^2-2n+n-2}{2}\right]}}{2} \\ \Rightarrow {}^{m}C_2=\frac{{\displaystyle \frac{n\left(n-1\right)}{2}\left[\frac{n\left(n-2\right)+1\left(n-2\right)}{2}\right]}}{2} \\ \Rightarrow {}^{m}C_2=\frac{{\displaystyle \frac{n\left(n-1\right)}{2}\frac{\left(n-2\right)\left(n+1\right)}{2}}}{2} \\ \Rightarrow {}^{m}C_2=\frac{\left(n+1\right)n\left(n-1\right)\left(n-2\right)}{8} \\ \Rightarrow {}^{m}C_2=\frac{\left(n+1\right)n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{8\left(n-3\right)!} \\ \Rightarrow {}^{m}C_2=\frac{3\left(n+1\right)!}{3\times 8\left[\left(n+1\right)-4\right]!} \\ \Rightarrow {}^{m}C_2=3·\frac{\left(n+1\right)!}{24\left[\left(n+1\right)-4\right]!} \\ \Rightarrow {}^{m}C_2=3·\frac{\left(n+1\right)!}{4!\times \left[\left(n+1\right)-4\right]!} \\ \Rightarrow {}^{m}C_2=3·{}^{n+1}C_4\left[\because {}^{n}C_m=\frac{n!}{m!\left(n-m\right)!}\right] \end{gathered}$$

Thus, ${}^{m}C_2$ is equal to $3·{}^{n+1}C_4$.

Hence, (C) is the correct option.