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Given natural numbers $x,y$, are there some identities between to $(x+y)!$, $x!$, $y!$, and some sum of "mixed" terms with $x$ and $y$? Essentially, is there a nice expansion of the terms if one were to expand,

$(x+y)!=(x+y)(x+y-1)\,\cdots\,(x+y-k)\,\cdots\,2\cdot1$

I can't seem to find anything relevant. I am open to generalizations to the gamma function if such identities exist in this more general context.

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    $\begingroup$ It would help if you explained why you needed these identities. $\endgroup$ – Mike Earnest Sep 6 '16 at 17:58
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    $\begingroup$ @MikeEarnest How the "why" shall be helpful? It's like if I asked you "what is the sum of the first $n$ natural numbers?" Would you need a reason why I need it, to answer me? Useless. Flagged. $\endgroup$ – Von Neumann Sep 6 '16 at 17:59
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    $\begingroup$ @FourierTransform The question "1+2+...+n=?" has one answer, so you don't need to know the application to answer it. The question "what identities involve x,y and (x+y)!"? has infinitely many answers, so you need further info to narrow them down and choose one. $\endgroup$ – Mike Earnest Sep 6 '16 at 18:04
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    $\begingroup$ @FourierTransform: Explaining why you need it is helpful to you. Telling the people trying to help you the context of your question lets them help you more usefully. Look at this thread on meta.SE. $\endgroup$ – Zev Chonoles Sep 6 '16 at 18:04
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    $\begingroup$ @FourierTransform I do agree with your point. If you look at questions I have asked here and at math overflow, they are typically well posed and thought out. I did not elaborate because I have not encountered a single identity involving $(x+y)!=$? While it is quite possible that I have overlooked some common identities, I still don't think there are many nice relations. I wonder if my reason is "I want to know of such identities" is really that unreasonable. $\endgroup$ – Joseph Zambrano Sep 7 '16 at 10:16
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Some possible developments are $$ \begin{gathered} (n + m)!\quad \left| {\,n,m \in \,\mathbb{N}\,\;} \right. = \left( {n + m} \right)^{\,\underline {\,n + m\,} } = \hfill \\ = \left( {n + m} \right)^{\,\underline {\,n\,} } m^{\,\underline {\,m\,} } = \left( {m + 1} \right)^{\,\overline {\,n\,} } m^{\,\underline {\,m\,} } = \hfill \\ = m!\sum\limits_{0\, \leqslant \,k\, \leqslant \,\min \left( {n,m} \right)} {\left( \begin{gathered} n \\ k \\ \end{gathered} \right)\;n^{\,\underline {\,n - k\,} } \;m^{\,\underline {\,k\,} } } = \hfill \\ = m!\,n!\sum\limits_{0\, \leqslant \,k\, \leqslant \,\min \left( {n,m} \right)} {\;\frac{{n^{\,\underline {\,n - k\,} } \;m^{\,\underline {\,k\,} } }} {{\left( {n - k} \right)!k!}}} = \hfill \\ = m!\,n!\sum\limits_{0\, \leqslant \,k\, \leqslant \,\min \left( {n,m} \right)} {\;\left( \begin{gathered} n \\ k \\ \end{gathered} \right)\left( \begin{gathered} m \\ k \\ \end{gathered} \right)} = \hfill \\ = m!\,n!\left( \begin{gathered} n + m \\ m \\ \end{gathered} \right) \hfill \\ \end{gathered} $$ with $$ \begin{gathered} n^{\,\underline {\,q\,} } = n\left( {n - 1} \right) \cdots \left( {n - q + 1} \right):\text{falling}\,\text{factorial} \hfill \\ n^{\,\overline {\,q\,} } = n\left( {n + 1} \right) \cdots \left( {n + q - 1} \right):\text{rising}\,\text{factorial} \hfill \\ \end{gathered} $$

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That's pretty broad... the only naturally useful fact I can think of is that $x!y!\mid (x+y)!$, and their quotient is the binomial coefficient $\binom{x+y}{x}=\binom{x+y}{y}$.

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  • $\begingroup$ Does the downvoter care to comment? $\endgroup$ – Zev Chonoles Sep 6 '16 at 17:54

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