Question

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

• A. ${}^{m+1}{C}_4$
• B. $2{}^{m+1}{C}_2$
• C. $3{}^{m+1}{C}_4$
• D. $3{}^{m+1}{C}_2$

Answer 1

The correct option is C $3{}^{m+1}{C}_4$

We know that,
$n={}^m{C}_2=\frac{m(m-1)}{2}$
Now,
${}^n{C}_2=\frac{n(n-1)}{2}\Rightarrow {}^n{C}_2=\frac{m(m-1)(\frac{m(m-1)}{2}-1)}{2\times 2}\Rightarrow {}^n{C}_2=\frac{m(m-1)(m^2-m-2)}{8}\Rightarrow {}^n{C}_2=\frac{m(m-1)(m+1)(m-2)}{8}\Rightarrow {}^n{C}_2=4!\times \frac{(m+1)m(m-1)(m-2)(m-3)!}{8\times 4!(m-3)!}\Rightarrow {}^n{C}_2=4!\times \frac{{}^{m+1}{C}_4}{8}\therefore {}^n{C}_2=3{}^{m+1}{C}_4$