Question

If ${{}^{10}P}_r=5040$ find the value of $r$.

Answer 1

Solution:-

${{}^{10}P}_r=5040$

As we know that,

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

$\therefore {{}^{10}P}_r=\frac{10!}{(10-r)!}$

$\Rightarrow \frac{10!}{(10-r)!}=5040$

$\Rightarrow \frac{10!}{(10-r)!}=7!(\because 7!=5040)$

$\Rightarrow (10-r)!=\frac{10!}{7!}$

$\Rightarrow (10-r)!=\frac{10\times 9\times 8\times 7!}{7!}$

$\Rightarrow (10-r)!=720$

$\Rightarrow (10-r)!=6!$

$\Rightarrow 10-r=6$

$\Rightarrow r=4$

Hence, $r=4$.