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 🙂
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 🙂