Question

If ${}^{10}{P}_r=604800$ and ${}^{10}{C}_r=120$, find r.

Answer 1

${}^{10}{P}_r=604800$

$\Rightarrow \frac{10!}{(10-r)!}=604800$----------1

${}^{10}{C}_r=120$

$\Rightarrow \frac{10!}{r!(10-r)!}=120$---------------------2

On dividing Equation 1 by Equation 2,

$\Rightarrow \frac{\frac{10!}{(10-r)!}}{\frac{10!}{r!(10-r)!}}=\frac{604800}{120}$

$\Rightarrow r!=\frac{604800}{120}$

$r!=5040$

$\therefore r=7$