Upper bound of binomial distribution

Question

I'm studying a proof and I'm wondering which binomial approximation could have been use to establish the following bound:

$$\cfrac{1}{2} {n \choose r} {n-r \choose r} \le \cfrac{n^{2r}}{2(r!)^2}$$

I get that:

$$\cfrac{1}{2} {n \choose r} {n-r \choose r} = \cfrac{n!}{2(r!)^2 (n-2r)!}$$

So I'm wondering about:

$$\cfrac{n!}{(n-2r)!} \le n^{2r}$$

Answer 1

You have $\frac{n!}{(n-2r)!}=(n-2r+1)\cdot(n-2r+2)]\cdot...\cdot n\le n^{2r}$ as it is a product of $2r$ numbers each of which is $\le$ than $n.$