# 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))

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

$\binom{n}{n/3} \approx \frac{1}{\sqrt{2n\pi(2/9)}(4/27)^{n/3}} = \frac{3(27/4)^{n/3}}{2\sqrt{n\pi}}$.

• If k > n, the binomial coefficient is zero. – marty cohen Apr 29 '15 at 2:28
• Not if n and k are integers. $\binom{n}{k} = n(n-1)...(n-k+1)/k! = 0$ if $k > n$. – marty cohen Apr 29 '15 at 2:43
• Stirling's formula doesn't know about integers and is only an approximation. – marty cohen Apr 30 '15 at 1:05
• I meant that one can use an asymptotic, large argument expansion of the Gamma function to derive Stirling's formula and can do so for non-integers. – Mark Viola Apr 30 '15 at 2:24

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}

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: $$\sqrt{\frac{a}{2\pi(a-1)n}}\cdot\left(\frac{a^a}{(a-1)^{a-1}}\right)^n\cdot\left(1+\frac{a-1-a^2}{12a(a-1)}\cdot\frac{1}{n}+O\left(\frac{1}{n^2}\right)\right)$$