What's the degree of $\binom{x}{k}$? Consider $\binom{x}{k}$, where $x$ is a positive integer variable and $0 \leq k \leq x$, where $k$ can be dependent on $x$. I'm interested in expanding $\binom{x}{k}$ in $x$.
For 'small' values of $k$, this is just a polynomial in $x$ of degree $k$. For example:
\begin{align*} \binom{x}{0} &= 1 &\text{ degree 0}\\
 \binom{x}{1} &= x &\text{ degree 1}\\
\binom{x}{2} &= \frac{x(x-1)}{2!} = \frac{1}{2}x^2 - \frac{1}{2} x &\text{ degree 2}\\
 \binom{x}{3} &= \frac{x(x-1)(x-2)}{3!} = \frac{1}{6} x^3 - \frac{1}{2} x^2 + \frac{1}{3} x &\text{ degree 3}\\
\end{align*}
However, as $k$ grows larger relative to $x$, this pattern no longer holds. As an extreme example, $\binom{x}{x-1} = x$ does not have degree $x-1$. Rather it has degree $1$, by the identity $\binom{x}{k} = \binom{x}{x-k}$ for all $k$. Similarly, $\binom{x}{x-2}$ has degree $2$.
But when $k$ is not extremely small or extremely large relative to $x$, $\binom{x}{k}$ is more complicated. For, say, $\binom{x}{\lfloor x/2 \rfloor}$, it's unclear what its degree should be, or even if the function $x \mapsto \binom{x}{\lfloor x/2 \rfloor}$ is a polynomial in $x$ at all. 
In short, my question is: Let $x \in \Bbb Z^+$, and let $k : \Bbb Z^+ \to \Bbb Z^{\geq 0}$ be a function of $x$ such that $0 \leq k(x) \leq x$ for all $x$. Consider the function $f: \Bbb Z^+ \to \Bbb Z^{\geq 0}$ given by: $f : x \mapsto \binom{x}{k(x)}$. For which functions $k$ is $f$ a polynomial? If it is not always a polynomial, what sort of function is $f$? 
What I have shown in my post is: When $k$ is a constant function, then $f$ is a polynomial. But there exist non-constant functions $k$ for which $f$ is a polynomial, for example $k(x) = x-1$. I'm wondering whether other functions, such as $k(x) = \lfloor x/2 \rfloor$, also produce polynomials. 
 A: Partial answer: the function $$ n \mapsto \binom{2n}{n} $$ is not a polynomial in $n$. By Stirling's approximation,
$$ \binom{2n}{n} = \frac{(2n)!}{(n!)^2} \sim \frac{\sqrt{4 \pi n} (2n/e)^{2n}}{2 \pi n (n/e)^{2n}} = \frac{4^n}{\sqrt{\pi n}} $$
which is exponential growth. In particular the function grows faster than any given polynomial in $n$.
A: See that for $k(x)<\frac x2$:
$$
 (\frac{x}{k(x)})^{k(x)}\leq\binom{x}{k(x)}\leq (\frac{xe}{k(x)})^{k(x)}.
$$
Now some examples:


*

*$k=px$ for $p\in(0,\frac 12)$.
$$
 (\frac{1}{p)})^{pn}\leq\binom{x}{k(x)}\leq (\frac{e}{p})^{np},
$$ 
which shows that the function behaves exponentially.

*$k$ is constant.
$$
 (\frac{x}{k})^{k}\leq\binom{x}{k(x)}\leq (\frac{xe}{k})^{k}.
$$
Then of course the function behaves as a polynomial.

In other words, by looking at :
$$
 (\frac{x}{k(x)})^{k(x)}\leq\binom{x}{k(x)}\leq (\frac{xe}{k(x)})^{k(x)}.
$$
one can say something about $f(x)$  by looking at $(\frac{x}{k(x)})^{k(x)}$.
To see whether asymptotically the function acts like a  polynomial, we can look at:
$$
K=\lim_{x\to\infty}\frac{\log\binom{x}{k(x)}}{\log x}.
$$
$K$ gives the degree of the asymptotic polynomial. 
For instance let's pick $k=p\log x$ for some positive constant $p$:
$$
 (\frac{x}{p\log(x)})^{p\log(x)}\leq\binom{x}{k(x)}\implies\\
 \lim_{x\to\infty}\frac{\log(\frac{x}{p\log(x)})^{p\log(x)} }{\log x}
\leq K=\lim_{x\to\infty}\frac{\log\binom{x}{k(x)}}{\log x}.
$$
But then the limit of the left hand side goes to infinity, showing that this will not be polynomial. As a matter of fact, the function behaves as $x^{p\log(x)}$ asymptotically. 
You can start to play with other choices of $k(x)$.
A: For a general $f(x)$, probably the best approach is to use the Integral representation of the Binomial
$$
\eqalign{
  & \left( \matrix{
  x \cr 
  f(x) \cr}  \right)\quad \left| \matrix{
  \;{\rm real }x \hfill \cr 
  \; - 1 < {\rm real }f(x) \hfill \cr}  \right.\;\quad  = {1 \over {2\,\pi }}\int_{\, - \,\pi \;}^{\,\pi \,} {e^{\, - \,i\,\,t\,f(x)} \left( {1 + e^{\,\,i\,\,t} } \right)^{\,x} \;d\,t}  =   \cr 
  &  = {1 \over {2\,\pi }}\int_{\, - \,\pi \;}^{\,\pi \,} {\sum\limits_{0\, \le \,j} {\left( \matrix{
  x \cr 
  j \cr}  \right)e^{\,\,i\,\,t\,\left( {j - f(x)} \right)} } \;d\,t}  = {1 \over {2\,\pi }}\sum\limits_{0\, \le \,j} {{{x^{\,\underline {\,j\,} } } \over {j!}}\int_{\, - \,\pi \;}^{\,\pi \,} {e^{\,\,i\,\,t\,\left( {j - f(x)} \right)} \;d\,t} }  \cr} 
$$
and from here to expand in terms of $x$, if $f(x)$ is analytic.
The simple case of $f(x)=x-m$ , with $m$ a positive integer is easily resolved via the gamma function, to give a polynomial
$$
\left( \matrix{
  x \cr 
  x - m \cr}  \right) = {{\Gamma (x + 1)} \over {\Gamma (m + 1)\;\Gamma (x - m + 1)}} = \left( \matrix{
  x \cr 
  m \cr}  \right) = {1 \over {m!}}x^{\,\underline {\,m\,} } 
$$
Also the case $f(x)=x/2$ can be easily approached, by using the duplication formula for Gamma, i.e.
$$
\left( \matrix{
  z \cr 
  z/2 \cr}  \right) = {{\Gamma \left( {z + 1} \right)} \over {\Gamma \left( {z/2 + 1} \right)^{\,2} }} = {{2^{\,\,z} } \over {\sqrt \pi  }}{{\Gamma \left( {z/2 + 1/2} \right)} \over {\Gamma \left( {z/2 + 1} \right)}} = {{2^{\,\,z - 1} } \over {z\sqrt \pi  }}{{\Gamma \left( {z/2 + 1/2} \right)} \over {\Gamma \left( {z/2} \right)}}
$$
(I have used $z$ to underline that it is valid also in the complex field)
and applying the Stirling's series
$$
\left( \matrix{
  z \cr 
  z/2 \cr}  \right) \propto {1 \over {2\sqrt {2e\pi } }}{1 \over {\sqrt z }}\left( {4{{z + 1} \over z}} \right)^{z/2} \quad \left| {\;z\, \to \,\infty ,\;\;\left| {\,\arg (z)\,} \right|} \right. < \pi 
$$
