5

So we know that $\large\sum_\limits{i=0}^t\dbinom{m}{i}\dbinom{n-m}{t-i}=\dbinom{n}{t}$ by a simple counting argument.

Now is there any bound on the quantity $\large\sum_\limits{i=0}^t(-1)^i\dbinom{m}{i}\dbinom{n-m}{t-i}$?

Can we show any non trivial upper bound on this quantity other than $\dbinom{n}{t}$?

JMP
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Vaas
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    isn't it rather $\sum_{i=0}^t\binom{m}{i}\binom{n-m}{t-i}=\binom{n}{t}$? – René Gy Feb 12 '16 at 18:42
  • Sorry my bad! ur right – Vaas Feb 13 '16 at 07:50
  • you should look at the method used in http://math.stackexchange.com/questions/1618417/prove-that-sum-k-0m-dbinomnk-dbinomn-km-k-2m-dbinomnm-for/1620874#1620874 and http://math.stackexchange.com/questions/745051/properties-of-a-sequence-of-sums-of-binomials/750615#750615 – reuns Feb 13 '16 at 08:09
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    since $\sum_{i=0}^t(-1)^i\binom{m}{i}\binom{n-m}{t-i}=\binom{n}{t} $ when $m=0$, any sharper bound should involve $n$, $t$ and $m$ – René Gy Feb 14 '16 at 17:43
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    It is the coefficient of $x^t$ in $(1-x)^m(1+x)^{n-m}$. In particular, when $n=2m$, it is zero when $t$ is odd. Not sure how that gives us any bound. – Thomas Andrews Feb 21 '16 at 00:52
  • This may be a formalized approach. Wolfram Alpha will give immediately $ {n-m \choose t} \quad {_2{F_1}}(-m, -t, -m+n-t+1, -1) $ and one may use properties of the hypergeometric function ${_2{F_1}}$ to arrive at bounds, e.g. use the Euler integral form. – Andreas Sep 19 '16 at 11:32

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What you ask for is the sign-alternating analogue of the Chu-Vandermonde convolution.

While I haven't proved that there is no closed formula, there is an interesting statement in the 1992 article by Andersen and Larsen, Combinatorial Summation Identities. The authors say that "the sign-alternating analogue [...] of the Chu-Vandermonde convolution is hardly known". For 3 special cases, they give the formulae by Kummer.

In his 2006 book "Summa Summarum", Larsen doesn't state new results either, so the 1992 quote may still hold.

Andreas
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