Question

If ${}^{2n}{P}_3=100{\times }^n{P}_2$, find $n$.

Answer 1

A] ${}^{2n}{P}_3=100{\times }^n{P}_2$

=$\frac{\left(2n\right)!}{(2n-3)!}=100\times \frac{n!}{(n-2)!}$

=$\frac{(2n-3)!(2n-2)(2n-1)2n}{(2n-3)!}=\frac{100\times (n-2)!(n-1)n}{(n-2)!}$

=2(2n-2)(2n-1)=100(n-1)

=$2(4n^2-6n+2)=100n-100$

=$8n^2-12n+4=100n-100$

=$8n^2-112n+104=0$

=$4n^2-56n+52=0$

=$2n^2-28n+26=0$

=$=n^2-14n+13=0$

= n (n-13) - 1 (n-13) = 0

n = 1, n = 13

$\therefore$ n = 13 [as n=1 not possible]