Question

If $^{32}{{P}_{6}}=k\,{{(}^{32}}{{C}_{6}}),$ then $k$ is equal to

• A. $6$
• B. $24$
• C. $120$
• D. $720$

Answer 1

Answer: (D)

$720$

Answer 2

Given, $^{32}{{P}_{e}}=k\,{{(}^{32}}{{C}_{6}})$
$\Rightarrow$ $\frac{32!}{(32-6)!}=k.\frac{32!}{6!(32-6)!}$ $\left[ \because \,{{\,}^{n}}{{P}_{r}}=\frac{n!}{(n-r)!}\,\,and{{\,}^{n}}{{C}_{r}}=\frac{n!}{r!(n-r)!} \right]$
$\Rightarrow$ $1=\frac{k}{6!}\Rightarrow k=6!$
$\Rightarrow$ $k=6\times 5\times 4\times 3\times 2\times 1=720$