Question

If ${}_n{C}_r=60$ and ${}^n{P}_r=360$, then find $r$?

• A. $5$
• B. $3$
• C. $6$
• D. Can't be determined

Answer 1

The correct option is B $3$

Given ${}_n{C}_r=60$ and ${}^n{P}_r=360$.
We also know that ${}_n{C}_r=\frac{n!}{(n-r)!\times r!}$ and ${}^n{P}_r=\frac{n!}{(n-r)!}$

On dividing ${}^n{P}_r$ with $r!$ we get,
$\frac{{}^n{P}_r}{r!}=\frac{n!}{(n-r)!}\times \frac{1}{r!}$

$\Rightarrow \frac{{}^n{P}_r}{r!}=\frac{n!}{(n-r)!r!}={}_n{C}_r$

$\Rightarrow \frac{{}^n{P}_r}{r!}={}_n{C}_r$

Now putting values of ${}^n{P}_r$ and ${}_n{C}_r$ in the above equation we get,
$\Rightarrow \frac{360}{r!}=60$

$\Rightarrow \frac{360}{60}=r!$

$\Rightarrow r!=6=3\times 2\times 1=3!$

$\Rightarrow r=3$