How to find limit involving combinatorics?

Question

I would like to evaluate the following limit:

$$\lim_{x \to \infty} \frac{x!}{m!(x-m)!}{5^{-x}},$$

where $0<m<x$.

Answer 1

$\dfrac{x!}{m!(x-m)!}0.2^x\lt\dfrac{x^m(x-m)!}{m!(x-m)!}0.2^x=\dfrac{x^m}{m!}0.2^x$

$0\le\lim_{x \to \infty} \dfrac{x!}{m!(x-m)!}0.2^x\le\lim_{x\to\infty}\dfrac{x^m}{m!}0.2^x=0$

Answer 2

Consider $$A_x=\frac{5^{-x} x!}{m! (x-m)!}$$ Take logarithms of both sides and use Stirling approximation for both factorials. This should lead to $$\log(A_x)=-x \log (5)+m \log \left({x}\right)+\log \left(\frac{1}{m!}\right)+O\left(\frac{1}{x}\right)$$ I am sure that you can take it from here.