Exponential growth of a combinations expression

Question

Assume $K,K\cdot a$ positive integers with $K\cdot a< K$. I have the expression $\binom{K}{K\cdot a}$ and the author says that it grows exponentially with $K$ when $a$ is fixed. Why is that?

Answer 1

Easiest way to explain this is probably via Stirling's approximation to the factorial.

Stirling's approximation is $$ n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\text{ as }n\to\infty. $$ Recalling that $$ \binom{K}{Ka}=\frac{K!}{(Ka)!(K(1-a))!}, $$ this implies $$ \begin{align*} \binom{K}{Ka}&\sim\sqrt{\frac{2\pi K}{4\pi^2KaK(1-a)}}\cdot\frac{\left(\frac{K}{e}\right)^K}{\left(\frac{Ka}{e}\right)^{Ka}\left(\frac{K(1-a)}{e}\right)^{K(1-a)}}\\ &=\sqrt{\frac{1}{2\pi Ka(1-a)}}\cdot\left(\frac{1}{a^a(1-a)^{1-a}}\right)^K\\ &=C\cdot\frac{1}{\sqrt{K}}\cdot D^K \end{align*} $$ with $C$ and $D$ constants (depending on $a$) defined in the obvious way.

Now, because $Ka<K$, we have $a\in(0,1)$. (We must have $a>0$; $a=0$ does not yield exponential growth here.) You can verify that this implies $1< D\leq 2$, so that this is an exponentially growing function.