Find the natural number $n>2$ such that $\frac{n!}{(n-1)!} + \frac{n!}{3!(n-3)!} = 2\frac{n!}{2!(n-2)!}$

Question

I'm unsure how I'm supposed to solve the equation:

$$\frac{n!}{(n-1)!} + \frac{n!}{3!(n-3)!} = 2\frac{n!}{2!(n-2)!} $$ given that $n>2.$

Answer 1

$n! = n(n-1)!$

So, start simplifying the first part of your equality.

$\frac{n!}{(n-1)!} = \frac{n(n-1)!}{(n-1)!} = n$.

Do you see how to continue?