WAEC Further Mathematics 2013 Paper

${}^nC_3 = \frac{n!}{(n-3)! \, 3!}$

${}^nP_2 = \frac{n!}{(n-2)!}$

$\frac{{}^nC_3}{{}^nP_2} = \frac{n!}{(n-3)!\,3!} \div \frac{n!}{(n-2)!}$

$\frac{n!}{(n-3)!\,3!} \times \frac{(n-2)!}{n!} = \frac{(n-2)!}{(n-3)!\,3!}$

Note that $(n-2)! = (n-2) \times (n-3)! = (n-2)(n-3)!$

$\frac{(n-2)(n-3)!}{(n-3)!\,3!} = 1$

$\frac{n-2}{3!} = 1 \implies n-2 = 6$

$n = 2+6 = 8$