Probability of random subsets of the same size intersecting $A$ and $B$ are subsets drawn uniformly from a set $C$, such that $|A|=|B|=a$ and $|C|=n$.
For what $a$ is the probability of $A\cap B \ne \varnothing$ equal to  $50\%$?
I figured that if $|A|=a$ and $|B|=b$ then the probability that there will be no intersection is 
$$\frac{(n-a)!(n-b)!}{n!(n-a-b)!}$$
and for this special case, 
$$\frac{(n-a)!(n-a)!}{n!(n-2a)!}$$
If we expand it, we get:
$$\prod_{i=0}^{a-1}{\left(1-\frac{a}{n-i}\right)}$$
and I wasn't able to go further from here to solve when is it equal to $0.5$
 A: If we use Stirling's Approximation, that is $n!\sim\sqrt{2\pi n}\frac{n^n}{e^n}$, we get that
$$
\frac{\binom{n-a}{a}}{\binom{n}{a}}\sim\frac{\left(1-\frac an\right)^{2n-2a+1}}{\left(1-\frac{2a}n\right)^{n-2a+1/2}}
$$
Using $\left(1+\frac xn\right)^n=e^x\left(1+O\left(\frac{x^2}n\right)\right)$, we get
$$
\begin{align}
\frac{\left(1-\frac an\right)^{2n-2a+1}}{\left(1-\frac{2a}n\right)^{n-2a+1/2}}
&=\frac{\left(1-\frac{2a}n+\frac{a^2}{n^2}\right)^{n-a+1/2}}{\left(1-\frac{2a}n\right)^{n-2a+1/2}}\\
&=\left[1+\frac{\frac{a^2}{n^2}}{1-\frac{2a}n}\right]^{\,n}\frac{\left(1-\frac{2a}n\right)^{2a}}{\left(1-\frac{2a}n+\frac{a^2}{n^2}\right)^a}\left[\frac{1-\frac{2a}n+\frac{a^2}{n^2}}{1-\frac{2a}n}\right]^{1/2}\\
&=\left[1+\frac{a^2}{n^2}+O\left(\frac{a^3}{n^3}\right)\right]^n\left(1-\frac{2a}n+O\left(\frac{a^2}{n^2}\right)\right)^a\left[1+O\left(\frac{a^2}{n^2}\right)\right]\\
&=e^{-a^2/n}\left(1+O\left(\frac{a^3}{n^2}\right)\right)
\end{align}
$$
Finding the $a$ so that this is approximately $\frac12$, we get
$$
a\sim\sqrt{n\log(2)}
$$
