# Simplifying a generating function in two variables with two binomial coefficients

I'm trying to to make the below expression simpler, and it would be great if it could be expressed as something like $(x+y)^k$.

$$\sum_{i=0}^k\binom{n+1}i\binom{m+1}{k-i}x^iy^{k-i}$$

The number $\binom{n+1}i$ is the coefficient of $x^i$ in $(1+x)^{n+1}$, and similarly $\binom{m+1}{k-i}$ is the coefficient of $y^{k-i}$ in $(1+y)^{m+1}$. The expression above is in fact the $k$th Chern class $c_k$ of the direct sum of two projective spaces ($CP^n$ and $CP^m$), but the problem may be viewed as completely separate from differential geometry.

I've tried to see if the coefficients match up with coefficients in expansions like $(1+x+y)^{n+m}$, but I can't see any immediate pattern. So, my question is, is it possible to simply the above expression?

• Something I notice right off is that the two-variable nature of this is pretty slim, since you can factor out an overall $y^k$ and define $z=x/y$ to get a generating function in $z$ alone. (Obviously the final expression will probably look more symmetric with $x$ and $y$, but for purposes of analysis I think $z$ is preferable.) – Semiclassical Jul 29 '14 at 17:41
• Got something from the answer below? – Did Aug 13 '14 at 12:01
• @Did not really, I mean that's the position that I started from, since the Chern class of a direct sum is the product of the Chern classes of the summands. I abandoned this question since I didn't get any other answers, but I guess yours is as close as I can get. – Jānis Lazovskis Aug 13 '14 at 12:14

This is the coefficient of $t^k$ in the polynomial $(1+tx)^{n+1}\cdot(1+ty)^{m+1}$.
• One can play a bit with something like rewriting this into a product of powers of $\dfrac{1+tx}{1+t y}$ and $(1+t x)(1+ty)$ which might be useful if $n$ and $m$ are far apart v. nearly the same. But beyond that... – Semiclassical Jul 29 '14 at 17:51