A) \[^{n+1}{{C}_{4}}\]
B) \[{{2}^{n+1}}{{C}_{4}}\]
C) \[{{3}^{n+1}}{{C}_{4}}\]
D) \[{{4}^{n+1}}{{C}_{4}}\]
Correct Answer: C
Solution :
\[m{{=}^{n}}{{C}_{2}}\] \[\Rightarrow \] \[m=\frac{n(n-1)}{2}\] Now, \[^{m}{{C}_{2}}=\frac{m\,(m-1)}{2}\] \[=\frac{\frac{n(n-1)}{2}\,\,\left[ \frac{n(n-1)}{2}-1 \right]}{2}\] \[=\frac{n\,(n-1)\,({{n}^{2}}-n-2)}{8}\] \[=\frac{n\,(n-1)\,(n+1)\,(n-2)}{8}\] \[=3.\frac{(n+1)\,n\,(n-1)(n-2)}{4.3.2.1}\] \[={{3}^{n+1}}\,{{C}_{4}}\]
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