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)

Given, ${}^{32}{P}_e=k{(}^{32}{C}_6)$
$\Rightarrow$ $\frac{32!}{(32-6)!}=k.\frac{32!}{6!(32-6)!}$ $\left[\because {}^n{P}_r=\frac{n!}{(n-r)!}and{}^n{C}_r=\frac{n!}{r!(n-r)!}\right]$
$\Rightarrow$ $1=\frac{k}{6!}\Rightarrow k=6!$
$\Rightarrow$ $k=6\times 5\times 4\times 3\times 2\times 1=720$