Question

Решите на множестве N
$frac{n!}{(n-4)!}= frac{22n!}{(n-3)!}$

Answer 1

$frac{n!}{(n-4)!}= frac{22n!}{(n-3)!}$
Сокращаем на n!
$frac{1}{(n-4)!}= frac{22}{(n-3)!} \ \ frac{(n-3)!}{(n-4)!}= 22 \ \ frac{1*2*3*...*(n*4)*(n-3)}{1*2*3*...*(n*4)} =22 \ \ n-3=22 \ \ n=25$