Question

${}^n{p}_r=3024$ and ${}^n{C}_r=126$ ,then $r$ is:

• A. 5
• B. 4
• C. 3
• D. 2
• E. 1

Answer 1

Answer: (B)

Given, ${}^n{P}_r=3024$
$\therefore$ $\frac{n!}{(n-r)!}=3024$ and
${}^n{C}_r=\frac{n!}{r!(n-r)!}$
$\Rightarrow$ $126=\frac{3024}{r!}$
$\Rightarrow$ $r!=24=4!$
$\Rightarrow$ $r=4$