$$\binom{n}{\frac{n}2}$$
I tried to do some rough estimates, but I didn't succeed at all well. Can somebody give a clue? Oh and i Forgot to mention, without Stirling formula.
One thought is to do fraction and get: $$\frac{\frac{n}{2}+1}{\frac{n}{2}}\cdot \frac{\frac{n}{2}+2}{n-1}\cdot ...\cdot \frac{\frac{n}{2}+1+2\left(\frac{n}{2}-1\right)}{\frac{n}{2}-\left(\frac{n}{2}-1\right)}\:=\:\left(1+\frac{1}{\frac{n}{2}}\right)\cdot ...\cdot \left(1+2\cdot \frac{n}{2}+1\right)$$
$$\theta \left(n\right)=\frac{n}{2}\cdot \left(1+\frac{2}{n}\right)\le \left(1+\frac{1}{\frac{n}{2}}\right)\cdot ...\cdot \left(1+2\cdot \frac{n}{2}+1\right)\le \frac{n}{2}\left(2+n\right)=\theta \left(n^2\right)$$
Is that correct?
