the asymptotic growth of n choose floor(n/2)

Using Stirling’s approximation, you get

n! = \sqrt{2n\pi}(n/e)^n

If you substitute it into $\choose{n}{n/2}$, you should eventually end up with

2^{n+1/2}/\sqrt{n\pi}

PS. you might want to check my math before you actually use the answer 🙂