Dixon's Identity states: $$ \sum_{k} (-1)^k\binom {a+b}{b+k}\binom{b+c}{c+k}\binom{c+a}{a+k} = \binom{a+b+c} {a,b,c}$$

A bit of history: The case $a=b=c$ was proved by Dixon in 1891 using trigonometric integrals. He proved the general case 11 years later using other analytical methods. In 1916, MacMahon proved his master theorem which gives another short analytical proof for the identity.

Using the WZ-method, Zeilberger gave a short proof in 1990 (with the aid of his computer). In 2003, Victor Guo gave a short proof using polynomials.

But what really interests me are the combinatorial interpretations.

  1. Zeilberger gives a proof using sign-reversing involutions in this lecture (starting from 39:30). Does it exist in some paper? I believe he credits it to Foata in this opinion.

  2. In this monograph by Foata, there seems to be a combinatorial proof of Dixon's identity and other related identities (pages 37 to 40). I wasn't able to understand the part in pages 37-38, where the following is derived combinatorially: $$\binom{b+c}{c+k}\binom{c+a}{a+k}\binom{a+b}{b+k}=\sum_{n\ge |k|} \binom{a+b+c-n}{a-n,b-n,c-n,n+k,n-k}$$ Where both sides seem to count directed graphs with the following adjacency matrix: $$\begin{bmatrix} 0 & c+k & b-k \\ c-k & 0 & a+k\\ b+k & a-k & 0 \end{bmatrix}$$ The LHS counts it directly, and the RHS counts it by decomposing by using 5 basic cycles: $1\to 2 \to 1$ (appearing $a-n$ times), $1\to 3 \to 1$ ($b-n$ times), $2\to 3 \to 2$ ($c-n$ times), $1\to 2 \to 3 \to 1$ ($n+k$ times), $1\to 3 \to 2 \to 1$ ($n-k$ times). But I don't really understand how they actually count it.

    Here's a picture of the relevant pages: Pages 37-38

  3. Is the paper "100 years of Dixon's identity" by James Ward (published in 1991 in the Irish Mathematical Society Bulletin") available online?

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    $\begingroup$ I can at least try to address the question about the five paths. They seem to be cycles of a simple graph, without loops or multiple edges. If that is the case, then your counterexample cannot occur since you have both $3 \rightarrow 2$ and $3 \rightarrow 1$. There are 3 cycles involving two vertices (12, 13, 23), and 2 involving all three (123, 132). $\endgroup$ – Brian Hopkins Mar 28 '13 at 21:21
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    $\begingroup$ To question 3: It is, since 2014. maths.tcd.ie/pub/ims/bull27/bull27_46-54.pdf $\endgroup$ – darij grinberg Aug 9 '15 at 12:12
  • $\begingroup$ Thanks, @darijgrinberg! It's a good article to display the diversity of techniques for proving combinatorial identites. $\endgroup$ – Ofir Aug 9 '15 at 15:03

Here are some remarks on the identity proof related to your second question. Hopefully they will make you understand everything, feel free to ask more about this proof.

Remark 1 Brian Hopkins’ comment is half correct and half incorrect. It is indeed true that loops are forbidden (this is indicated by the “$i\neq j$” statement on the second line of p.37). On the other hand, multiple edges are definitely needed (and allowed) : otherwise, you could hardly multiply two cycles together and the whole proof would be nonsensical.

Remark 2 Your cycle is not a counterexample, as it can be decomposed as $[123][132]$ or $[132][123]$, depending on how you number your multiple edges. It is very important to remember that the multiple edges are NUMBERED, this is what allows us to make the product of cycles to be non-commutative (see the definition of the product on the middle of p.15).

Remark 3 The integers $a,b,c,n,k$ in formulas (1),(2),(3) on the bottom of page 37 are uniquely defined by $$ a=\varepsilon_3+\frac{\varepsilon_4+\varepsilon_5}{2}, \ b=\varepsilon_2+\frac{\varepsilon_4+\varepsilon_5}{2}, \ c=\varepsilon_1+\frac{\varepsilon_4+\varepsilon_5}{2}, \ n=\frac{\varepsilon_4+\varepsilon_5}{2}, \ k=\frac{\varepsilon_4-\varepsilon_5}{2} $$

Remark 4 There is a typo in the formula for matrix $N$ on the top of p.38 : the $(3,2)$ coefficient should be $a+k$ (not $b-k$ as it is printed).

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  • $\begingroup$ Thank you for your remarks. Regarding the 2nd remark: I indeed missed the fact that the edges are numbered. Can you please explain why 1→2→3→2→3→1 is decomposed into [123][132] and not into [123][23]? I am still missing something, since my example contains 5 edges and each of [123],[132] contains 3 edges. $\endgroup$ – Ofir Mar 30 '13 at 14:32
  • $\begingroup$ @Ofir : ah, I mistyped. The cycle $(1,2,3,2,3,1)$ would be decomposed as $[123][23]$ or $[23][123]$, depending on the numbering. $\endgroup$ – Ewan Delanoy Mar 30 '13 at 14:36
  • $\begingroup$ Great, I think I've understood it completely now. For example, the graph $1\to 2 \to 3 \to 1 \to 3 \to 2 \to 1$ can be decomposed in 8 ways: [1,2,3][3,2,1] (and in the other order) and [1,2][2,3][3,1] (and other 5 orders). Thank you for helping me understand this beautiful proof. I will wait another day (in case someone gives an answer to the other questions) before I give you the bounty. $\endgroup$ – Ofir Mar 30 '13 at 15:11
  • $\begingroup$ Congrats on getting the bounty :) I have some more questions, if you don't mind. 1. I understand how to turn a cycle decomposition into a graph (you just start numbering the edges in the first cycle, and then in the second, and so on). But I am not sure how you turn a graph into cycles - once you choose a node, it is simple (you start walking by choosing the smallest edge each time). But how do you choose the node? 2. Do you read French or did you use some translation tool? $\endgroup$ – Ofir Mar 31 '13 at 7:44
  • $\begingroup$ 1) Actually, only a circuit can be decomposed into cycles. Proposition 3.4. on page 17 describes the algorithm : there is always a cycle made only of edges labelled 1, so you can start with this cycle. 2) I do read French. $\endgroup$ – Ewan Delanoy Mar 31 '13 at 7:52

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