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Let $a>0, b>0,m>0$

$H(t)=\sum\limits_{k=0}^{\infty} {k \choose a}{m-k \choose b}t^k$

So what is the closed form of $H(t)$?

What I know currently is:

$\sum\limits_{0 \le k \le t} {t-k \choose r} {k \choose s} = {t+1 \choose r+s+1}$

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    $\displaystyle P_{m,a,b}(t)=\sum_{k=a}^{m-b}\binom{k}{a}\binom{m-k}{b}t^k$ is simply$a$polynomial with each coefficient given in closed form. As such, it is already in closed form. Is there some property of this polynomial that you suspect or would like to see? If so, please ask about that.2011-10-15

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$ H(t)=-\frac1{4\pi^2}\oint_{|z|=1}\oint_{|w|=1}\frac{(1+w)^{m+1}-(1+z)^{m+1}t^{m+1}}{1+w-(1+z)t}\,\frac{\mathrm dw}{w^{b+1}}\,\frac{\mathrm dz}{z^{a+1}} $

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    @robjohn, this answer **WAS** tongue-in-cheek, more precisely meant to be a kind of psychological preparation for my comment on the question. Could not say whether the intended message got through or not... Oh you, sweet mysteries of written communication... :-)2011-10-16