How can a binomial coefficient can be approximated by using Stirling's formula? I've met some difficulties with such question:
How can we approximate a binomial coefficient by using a Stirling's factorial approximation.
I've evaluate a little bit and got this
How can I transform the right side of this equation for getting estimation like (1 + ?/n + O(1/n^2))
 A: Another case is when
$k$ is a constant times $n$.
Let
$k = a n $
where $0 < a < 1$,
so $k/n = a$.
Then
$\begin{align}
\binom{n}{k}
&=\frac{n!}{k!(n-k)!}\\\\
&\approx. \frac{\sqrt{2\pi n}(\frac{n}{e})^n} {\sqrt{2\pi k}(\frac{k}{e})^k \sqrt{2\pi (n-k)}(\frac{n-k}{e})^{n-k}}\\\\
&=\frac{1}{\sqrt{2\pi k}(k/n)^k \sqrt{1-(k/n)}(1-(k/n))^{n-k}}\\
&=\frac{1}{\sqrt{2\pi k(1-k/n)}a^{an} (1-a)^{n-an}}\\
&=\frac{1}{\sqrt{2\pi n a(1-a)}(a^a(1-a)^{1-a})^n}\\
\end{align}
$
For example,
if $a = 1/2$,
since
$a(1-a) = 1/4$
and
$a^a (1-a)^{1-a}
=(1/2)^{1/2}(1/2)^{1/2}
=1/2
$,
this becomes
$\binom{n}{n/2}
\approx \frac{1}{\sqrt{2n\pi/4}(1/2)^n}
= \frac{2^n}{\sqrt{n\pi/2}}
$.
For another example,
if $a = 1/3$,
since
$a(1-a) = 2/9$
and
$a^a (1-a)^{1-a}
=(1/3)^{1/3}(2/3)^{2/3}
=\frac{(2^2)^{1/3}}{3}
=\frac{4^{1/3}}{3}
=(4/27)^{1/3}
$,
this becomes
$\binom{n}{n/3}
\approx \frac{1}{\sqrt{2n\pi(2/9)}(4/27)^{n/3}}
= \frac{3(27/4)^{n/3}}{2\sqrt{n\pi}}
$.
A: Stirling's approximation is given by $n!\approx. \sqrt{2\pi n}(n/e)^n$.  Thus, for large $n$, and $n>>k$, we have 
$$\begin{align}
\binom{n}{k}&=\frac{n!}{k!(n-k)!}\\\\
&\approx. \frac{\sqrt{2\pi n}(\frac{n}{e})^n}{k!\sqrt{2\pi (n-k)}(\frac{n-k}{e})^{n-k}}\\\\
&=\frac{n^ne^{-k}}{k!\sqrt{1-(k/n)}n^{n-k}(1-(k/n))^{n-k}}\\\\
&\approx. \frac{n^k}{k!(1-(k/n))^{-k+1/2}}\\\\
&=\frac{n^k}{k!}\,\left(1-\frac{k}{n}\right)^{k-1/2}
\end{align}$$
A: Thank you all for your answers, but I mentioned that I have to use slightly different factorial approximation:
$$
n!=\sqrt{2\pi n}\cdot(\frac{n}{e})^n\cdot\left(1+\frac{1}{12n}+\frac{1}{288n^2}+O(\frac{1}{n^3})\right)
$$
So, by using this approximation it's necessary to obtain approximation for the binomial coefficient $\binom{an}{n}$. Sorry that I defined the task conditions rudely:
$a$ - is a constant and $n\rightarrow \infty$. 
So, after doing some conversions I've got this one:
$$\sqrt{\frac{a}{2\pi(a-1)n}}\cdot\left(\frac{a^a}{(a-1)^{a-1}}\right)^n\cdot\frac{1+\frac{1}{12an}+\frac{1}{288(an)^2}}{\left({1+\frac{1}{12n}+\frac{1}{288n^2}}\right)\cdot\left(1+\frac{1}{12(a-1)n}+\frac{1}{288((a-1)n)^2}\right)}$$
Then let's transform the right side of the expression:
$$\frac{1+\frac{1}{12an}+\frac{1}{288(an)^2}}{\left({1+\frac{1}{12n}+\frac{1}{288n^2}}\right)\cdot\left(1+\frac{1}{12(a-1)n}+\frac{1}{288((a-1)n)^2}\right)}=\\=\displaystyle{\frac{1+\frac{1}{12an}+O(\frac{1}{n^2})}{1+\frac{a}{12(a-1)n}+O(\frac{1}{n^2})}}$$
Finally, we get:
$$\displaystyle{1+X=\frac{1+\frac{1}{12an}}{1+\frac{a}{12(a-1)n}}}\rightarrow X=\frac{a-1-a^2}{a(12(a-1)n+a)}$$So, the right side of our approximation can be submitted as:
$$1+\frac{a-1-a^2}{12a(a-1)}\cdot\frac{1}{n}+O\left(\frac{1}{n^2}\right)$$
And the whole answer is:
$$\sqrt{\frac{a}{2\pi(a-1)n}}\cdot\left(\frac{a^a}{(a-1)^{a-1}}\right)^n\cdot\left(1+\frac{a-1-a^2}{12a(a-1)}\cdot\frac{1}{n}+O\left(\frac{1}{n^2}\right)\right)$$
