# Is $\binom{n}{k}$ always considered polynomial?

I'm trying to understand the growth of the term $\binom{n}{k}$ - I saw here a proof that $\binom{n}{k} = O(n^k)$. However, if $k$ is quite large (say $k=n$) then this term is not polynomial. I know that $\binom{n}{k} = \binom{n}{n-k}$ and hence we can deduce that $\binom{n}{k} = O(n^{\min \{k, n-k \}}$.

My problem regards to the worst case, that is when $k=\frac{n}{2}$ (let's assume that $n$ is even). In this case $k = n-k$ and we get $\binom{n}{k} = O(n^{\frac{n}{2}})$, which seems very-not-polynomial to me...

Am I missing something? Or that is just my basic assumptions about the binomial coefficient being a polynomial is simply wrong?

• It depends on what is a variable. In this case I assume $n$ is our variable and $k$ is constant, so saying $n=k$ does not make sense. – I was suspended for talking Apr 15 '18 at 12:50
• The variable is indeed $n$. My question actually regards for a theoretical algorithm for the vertex cover problem, so I need to take care of the worst case with regards to $k$ (clearly there's a graph with minimal VC of size $k=n$). – noamgot Apr 15 '18 at 12:52
• Yes, $\binom{2n}{n}=\frac{(2n)!}{(n!)^2}\sim\frac{2^{2n}}{\sqrt{\pi n}}$, by Stirling's approximation, is not polynomial. – user552631 Apr 15 '18 at 12:59
• You can get an approximation to $\binom {2n}n$ or $\binom {2n+1}n$ using Stirling's approximation to the factorials. Note that this is not of the form $\binom nk$ with $k$ fixed so is not polynomial in form. – Mark Bennet Apr 15 '18 at 13:00
• Did you read the comments on the question you link to? The very first one explains that the binomial is not necessarily polynomial when $k$ is a function of $n$ (which it is, in your case: $k=n/2$). – David K Apr 15 '18 at 13:29

For $n$ and $k$ large, it is helpful to think of this number in terms of Stirling's approximation.

Recall that

$${n \choose k}= \frac{n!}{k! (n-k)!}$$

and

$$n!\approx \sqrt{2\pi}\cdot \frac{n^{n+1/2}}{ e^{n} }.$$

Hence

$${n \choose k}\approx \frac{1}{\sqrt{2\pi}} \sqrt{\frac{n}{k(n- k)}}\left(\frac{n}{k}\right)^{k} \left(\frac{n}{n-k}\right)^{n-k}.$$

This approximation is good when $n$, $k$ and $n-k$ are large. Like people have remarked in the comments, if something is polynomial depends on what you consider variables and what constants.

If both $n$ and $k$ are variables then this thing is not polynomial.

If $n$ is constant, then there are finitely many possibilities for $k$, so the thing is clearly polynomial (bound by a constant).

If $n$ is considered variable but $k$ constant, then observe that $$f(k)={n\choose k}=\frac{1}{k!}n\cdot (n-1)\cdot ...\cdot (n-k+1),$$ which is a polynomial in $n$.

For $0<\alpha<1$, we have $$\binom{n}{n\alpha}^{-1} = (n+1)\int_0^1 x^{\alpha n} (1-x)^{(1-\alpha)n} \, dx$$ (this is a consequence of the more general relationship $\binom{a+b}{a}^{-1} = (1+a+b)\int_0^1 x^a (1-x)^b \, dx$).

We can use Laplace's Method to find an asymptotic for this integral, whence $$\binom{n}{n\alpha} \sim \frac{1}{\sqrt{2\pi \alpha(1-\alpha)}}\frac{((1-\alpha)^{-1+\alpha}\alpha^{-\alpha})^{n}}{\sqrt{n}},$$ which is indeed not polynomial.