A) \[{{3}^{n}}{{C}_{4}}\]
B) \[^{n+1}{{C}_{4}}\]
C) \[{{3}^{n+1}}{{C}_{4}}\]
D) \[{{3}^{n+1}}{{C}_{3}}\]
E) \[{{3}^{n+1}}{{C}_{2}}\]
Correct Answer: C
Solution :
\[\because \]\[m{{=}^{n}}{{C}_{2}}=\frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2}\] Now, \[^{m}{{C}_{2}}=\frac{m!}{2!(m-2)!}=\frac{m(m-1)}{2}\] \[=\frac{\underset{2}{\mathop{n(n-1)}}\,.\left( \begin{align} & {{n}^{2}}-n-2 \\ & \,\,\,\,\,\,\,\,\,2 \\ \end{align} \right)}{2}\] \[=\frac{(n+1)n(n-1)(n-2)}{8}\] \[={{3.}^{n+1}}{{C}_{4}}\]
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