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For partial binomial sums such as $\sum_{k\le\Delta} \binom{n}{k}$ we don't tend to have closed forms. However we still know asymptotic expansions that are easy to work with.

Can we do something similar in two dimensions, for a sum like the following guarded by a number of linear inequalities?

$$\sum_{\substack{0\ \le\ i,\ j\\i+j\ \le\ \Delta\\r-i+j\ \le\ \Delta}} \binom{n}{i}\binom{m}{j}$$

For a start we could consider triangles, and then compose an answer, but that doesn't seem a lot easier. It doesn't even seem easy to identify which term might be the largest of the sum. Any ideas?

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What about working with continuous approximation ? Replacing the binomial by Gaussians, and the sum by integrals ? The integral of a multidimensional Gaussian is reasonably easy to compute over rectangle domains (product of $erf$), and you may get the simplex using a symmetry argument.

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  • $\begingroup$ I see how I may split the product when integrating over a rectangle, but how about over a triangle? Like $x+y\le K$? $\endgroup$ – Thomas Ahle Feb 11 '15 at 22:09
  • $\begingroup$ I may have overlooked the conditions... I was thinking that a triangle is just half of a rectangle, but this does not gives a solution, unless the rectangle has the same center as your Gaussian... $\endgroup$ – LeoDucas Feb 11 '15 at 22:14

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