First note that $\binom{m}{2} = \frac{m(m-1)}{2}$ and use that to get rid of the nested binomials. Also, the floor function will not (here) make any difference, so ignore it.

Then convert all binomials to their $\Gamma$ equivalents, and use Stirling's formulas for each term. The next step is the messiest, as you'll have a lot of arithmetic to perform on the result, which will give you the result.

This is sufficiently mechanical that, using Maple, I can quickly derive that
$$ \frac{e^{-2kc}}{k!} + \frac{-\frac{1}{2}e^{-2kc}((4c+k+1)\ln{n}+2kc+4c^2+\ln^2{n}-1+2c+k)}{n (k-1)!}+O(n^{-2})$$

Of course, that second term might not be quite right, since the previously ignored floor function might here contribute, I have not checked that.