<s>[INST] find set of points for lots of triangulations I should find a set of $n$ points $Q$ in a plane, so that $t(Q)$ (the number of possible triangulations) the following equation holds: 
$$t(Q) \ge 2^{n-2\sqrt{n}}$$

I solved the problem using the vertex points of a convex polygon. Wikipedia says, that $t(Q) = C_{n-2}$ and by induction I can show, that $C_{n-2} \ge  2^{n-2\sqrt{n}}$. 
I think, that this approach is way to complex. There must be a better solution, because $C_{n-2} \gg  2^{n-2\sqrt{n}}$ for big $n$. 
Any ideas?
 [/INST] How complex is too complex?
The central binomial coefficient $\displaystyle \binom{2n}{n}$ is bigger than $\displaystyle \frac{4^n}{2n+1}$, so 
$\displaystyle t(Q) = C_{n-2} = \frac{1}{n-1} \binom{2(n-2)}{n-2} \geq \frac{1}{n-1} \frac{4^{n-2}}{2n - 3} = \frac{2^{n+2\sqrt{n} - 4}}{(n-1)(2n-3)} 2^{n-2\sqrt{n}}$
Since $2^{n+2\sqrt{n} - 4} \geq (n-1)(2n-3)$ for $n > 3$, and $C_{3-2} = 1 > 2^{3 - 2\sqrt{3}}$ we have $t(Q) \geq 2^{n-2\sqrt{n}}$
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