A) \[^{m+1}{{C}_{4}}\]
B) \[^{m-1}{{C}_{4}}\]
C) \[3\,.{{\ }^{m+2}}{{C}_{4}}\]
D) \[3\ .{{\ }^{m+1}}{{C}_{4}}\]
Correct Answer: D
Solution :
\[\alpha {{=}^{m}}{{C}_{2}}\Rightarrow \alpha =\frac{m(m-1)}{2}\] \[\therefore \]\[^{\alpha }{{C}_{2}}{{=}^{m(m-1)/2}}{{C}_{2}}=\frac{1}{2}.\frac{m(m-1)}{2}\left\{ \frac{m(m-1)}{2}-1 \right\}\] \[=\frac{1}{8}m(m-1)(m-2)(m+1)\] \[=\frac{1}{8}(m+1)\ m(m-1)(m-2)=3\ .{{\ }^{m+1}}{{C}_{4}}\]
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