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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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    What makes you think that there is a closed form? What do you know about hypergeometric functions?2011-10-12
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    I don't seem to be able to convince the site to interpret correctly a link into W|A... but do enter the text «sum[Binomial[k,a] Binomial[m-k,b] t^k, {k, 0, infinity}]» into Wolfram Alpha.2011-10-12
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    As always, you could turn all the binomial coefficients into Pochhammer symbols to obtain the hypergeometric representation...2011-10-12
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    Fan Zhang, as I observed above your series sums to a well-known hypergeometric function. Unless you add extra conditions on $a$, $b$ and $m$, there is really nothing more to be said, and anyone will tell you that the expression $\binom{0}{a} \binom{m}{b} \, _3F_2(1,1,b-m;1-a,-m;t)$ which Wolfram Alpha returns *is* the closed form.2011-10-15
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    @MarianoSuárez-Alvarez: the problem with Alpha links is usually the special characters. If you escape them with % followed by the ASCII code it works. The site converts most of them for you, but not all.2011-10-15
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    @Ross: in fact, I urlencoded the whole thing and SE seemed to think it was too long, and linkified only a part2011-10-15
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    FanZhang, I suggest you read and ponder [this page](http://en.wikipedia.org/wiki/Closed-form_expression). It might help you get the point (made by others in the comments) that very few functions can be written with the functions you know, and that $H$ is **not** one of them (for general values of the parameters $a$, $b$ and $m$).2011-10-15
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    $\binom{-n}{k}$ is not $0$ for $n>0$. Do you intend your sum to actually be $\displaystyle\sum_{k=a}^{m-b}\binom{k}{a}\binom{m-k}{b}t^k$?2011-10-15
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    @robjohn I forget that {-n choose k} is not 0 for $n>0$, actually I am looking for coefficient that upper item of binomial is $\ge$ the lower one.2011-10-15
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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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    [Since we're bringing out contour integrals anyway...](http://dlmf.nist.gov/16.5) :D2011-10-15
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    http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/ lists quite a few representations (it says there are 40964 formulas, but the immense majority are evaluations)2011-10-15
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    And, maybe more old fashioned but still useful, [Abramowitz and Stegun](http://people.math.sfu.ca/~cbm/aands/).2011-10-15
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    Nice simplification :-)2011-10-15
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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