I try to find an approximation for the series $S = \sum_{i=0}^n {{n \choose 2} \choose i}$ (A066383 on OEIS) of the form $S \sim \mathcal{O}(2^{\alpha \cdot N^2})$ for $n \to \infty$.
I tried the following, but did not succeed.
In my opinion, the largest term of $S$ is the one, when $i=N$ (since $i$ does not reach $\frac{{N \choose 2}!}{2}$): $$\frac{{N \choose 2}!}{N! \cdot ({N \choose 2} -N)!}$$
I then write $$e^{\ln \bigl( \frac{{N \choose 2}!}{N! \cdot ({N \choose 2} -N)!} \bigr)}$$ and use sterlings approximation $\ln x! \approx x \ln x$: $$e^{\frac{N(N-1)}{2} \cdot \ln (\frac{N(N-1)}{2}) - N \ln N - \bigl[ (\frac{N(N-1)}{2} -N) \bigr] \cdot \ln \bigl( \frac{N(N-1)}{2} -N \bigr)}$$ and at this point I am stuck and would be grateful for any advice.
Note: I read, that $S = 2^{1/2 (N - 1) N} - {1/2 (N - 1) N \choose N + 1} 2F1(1, -N^2/2 + (3 N)/2 + 1, N + 2, -1)$, where $2F1(a,b;c;x)$ is the hypergeometric function. But I have no experience with the hypergeometric function and therefore don't know, how the second term would scale in this case.
Thank you in advance