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In a book I have met one interesting equation (without reference): $$\frac{m!}{n!}\sum_{i=0}^n(-1)^i{n\choose{i}}{x+m+n-i\choose{m}}=\begin{cases} x+n+1,\, if \,m=n+1 \\ 1,\, if \,m=n \\ 0,\,if\, m<n. \end{cases}$$ First question is as follows: is there a simple answer for this sum for $m>n$? The second question: how can I find a name of the author of this equation (to make a reference in a paper).

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    $\begingroup$ It might make sense to mention the book where you saw it. $\endgroup$
    – user9072
    Commented May 30, 2014 at 12:28

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The identity

$$ \sum_{i=0}^n (-1)^i \binom{n}{i} \binom{y-i}{a} = \binom{y-n}{a-n} $$

is valid for all $y \in \mathbb{C}$ and $a, n \in \mathbb{N}_0$, provided the right-hand side is interpreted as $0$ when $n > a$. (Hint for proof: $\sum_{i=0}^n (-1)^i \binom{n}{i}$ is the $n$th iterated difference operator.) Hence

$$ \sum_{i=0}^n (-1)^i \binom{n}{i} \binom{x+m+n-i}{m} = \binom{x+m}{m-n}. $$

So it looks like the scaling factor $m!/n!$ at the start of your formula should be removed. Then the left-hand side is $x+n+1$ if $m = n+1$, as you stated, and generally if $m = n+r$ then the sum is $\binom{x+n+r}{r}$.

One reference is Concrete Mathematics, Equation (5.24): take $m=0$ and make a small rearrangement.

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