Question

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

• A. ${{}^{m+1}C}_4$
• B. ${{}^{m-1}C}_4$
• C. ${{}^{m+2}C}_4$
• D. none of these

Answer 1

The correct option is D none of these

We have $n={{}^{m}c}_2=\frac{m(m-1)}{2}$

${{}^{n}c}_2=\frac{n(n-1)}{2}$

${\Rightarrow }^n{C}_2=\frac{m(m-1)}{4}\left[\frac{m(m-1)}{2}-1\right]=\frac{1}{8}m(m-1)(m^2-m-2)$

${\Rightarrow }^n{C}_2=\frac{1}{8}m(m-1)(m-2)(m+1)$

${\Rightarrow }^n{C}_2=3\times \frac{1}{24}(m+1)m(m-1)(m-2)=3\times {{}^{m+3}c}_4$