A) \[6\]
B) \[24\]
C) \[120\]
D) \[720\]
Correct Answer: D
Solution :
Given, \[^{32}{{P}_{e}}=k\,{{(}^{32}}{{C}_{6}})\] \[\Rightarrow \] \[\frac{32!}{(32-6)!}=k.\frac{32!}{6!(32-6)!}\] \[\left[ \because \,{{\,}^{n}}{{P}_{r}}=\frac{n!}{(n-r)!}\,\,and{{\,}^{n}}{{C}_{r}}=\frac{n!}{r!(n-r)!} \right]\] \[\Rightarrow \] \[1=\frac{k}{6!}\Rightarrow k=6!\] \[\Rightarrow \] \[k=6\times 5\times 4\times 3\times 2\times 1=720\]
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