Compare $(n!)^2$ and $p!(2n-p)!$ with $0 \le p \le 2n$

Question

Here is the full exercise:

They told us to check $\forall n \in \Bbb N$

$$\left(\frac{n!}{(n-1)!}\right)^2$$ and $$\frac{n!}{2(n-2)!} +\frac{n+1!}{2(n-1)!}$$

Then they told us to compare $(n!)^2$ and $p!(2n-p)!$ for $0 \le p \le 2n$

I really didn't get any idea of how does this factorial works, is anyone here have an idea or a well-detailed solution?

Answer 1

We have that

$$\frac{(n!)^2}{p!(2n-p)!}=\frac{(2n)!}{p!(2n-p)!}\frac{(n!)^2}{(2n)!}=\frac{\binom{2n}{p}}{\binom{2n}{n}}\le 1$$

therefore $(n!)^2\le p!(2n-p)!$.