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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?

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    By $Ka$ do you mean the product $K\times a$ or $K_a$, a distinct integer from $K$? – 2017-01-16
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    @Mark Thanks, I have made an edit to clarify that it's actually a product – 2017-01-16
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    Are both $K$ and $Ka$ supposed to be positive? – 2017-01-16
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    @Mark Yes, they are – 2017-01-16
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    what about $a=1$... – 2017-01-16
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    I don't think $a$ has to be an integer (so as an example, $K = 8, a = 2$ might work). – 2017-01-16
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    @Rustyn I guess I'll have to make a lot of edits. Please see above. Let's stick to the case of strict inequality – 2017-01-16
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    @Mark First, I did not say that $a$ has to be an integer, $K\cdot a$ has to be. Second, in your example you set $a$ as an integer. – 2017-01-16

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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 $Ka0$; $a=0$ does not yield exponential growth here.) You can verify that this implies $1< D\leq 2$, so that this is an exponentially growing function.

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    Thanks, can you give me a hint on the proof of $D\geq2$? – 2017-01-16
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    That's equivalent to showing that $a^a(1-a)^{1-a}\leq\frac{1}{2}$. To that end: take the logarithm to get $a\ln(a)+(1-a)\ln(1-a)$. You can find the maximum of this by differentiation; then, you can finish using monotonicity of the exponential. – 2017-01-16
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    Hmm... hold on a sec. I've made a mistake here. $a^a(1-a)^{1-a}$ has a MINIMUM of $\frac{1}{2}$. One sec. – 2017-01-16
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    Alright, fixed. And you can show it by showing the minimum of $\frac{1}{2}$, as described above, combined with $a^a(1-a)^{1-a}<1$. – 2017-01-16
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    OK, I have made some calculations here. So, for the first part $D\leq2$ we have $\vartheta /\vartheta a(a\mathrm{ln}a+(1-a)\mathrm{ln}(1-a))=0\Rightarrow ln(a/(1-a))=0\Rightarrow a=1/2$ which yields only the maximum of $D$ or equivalently the minimum of $a^a(1-a)^{1-a}$. How can I get the other bound? – 2017-01-16
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    Just note that $0$a^a\in(0,1)$ and $(1-a)^{1-a}\in(0,1)$. – 2017-01-16