The question says :
Use Stirling's approximation to estimate $\binom{m+n}{n}$ when $m$ and $n$ are both large.
I did till following but not sure what's next step to do.
(Stirling's approximation is $n!$ ~ $\sqrt{2\pi n}(\frac{n}{e})^n$) .
So, $\binom{m+n}{n}=\frac{(m+n)!}{(m!)(n!)}$ ~ $\frac{\sqrt{2\pi(m+n)}(\frac{m+n}{e})^{m+n}}{\sqrt{2\pi m}(\frac{m}{e})^{m}\sqrt{2\pi n}(\frac{n}{e})^n}=\sqrt{\frac{1}{2\pi m}+\frac{1}{2\pi n}}(1+\frac{n}{m})^m(1+ \frac{m}{n})^n$.
What should be the next step. Thanks.
$$A=\binom{m+n}{n}=\frac{(m+n)!}{m!\,n!}$$ Take logarithms $$\log(A)=\log \Big[\binom{m+n}{n}\Big]=\log((m+n)!)-\log(m!)-\log(n!)$$
Now use Stirling approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \log ( p)+\frac{1}{2} \log (2 \pi )+\frac{1}{12 p}+O\left(\frac{1}{p^2}\right)$$ Apply it three times ignoring the higher order terms. Now, say that $m > n$ and continue with Taylor expansion for large $m$. This should give $$\log(A)=n \log \left(\frac{m}{n}\right)+ \left(n-\frac{1}{12 n}-\frac{1}{2} \log (2 \pi n)\right)+\frac{n (n+1)}{2 m}+O\left(\frac{1}{m^2}\right)$$
Use it for $m=100$ and $n=99$. The above twice truncated expansion would give $\log(A)=129.86$ while the exact value is $\log(A)=135.06$.