Prove $\sum_k{{a+b \choose a+k}{b+c \choose b+k}{c+a \choose c+k}(-1)^k}=\dfrac{(a+b+c)!}{a!b!c!}$

Asked Nov 26 '23 at 18:27

Active Nov 26 '23 at 18:42

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How can I prove this: $$\sum_k{{a+b \choose a+k}{b+c \choose b+k}{c+a \choose c+k}(-1)^k}=\dfrac{(a+b+c)!}{a!b!c!}$$

I know I should avoid no clue questions, but really I have no idea about this one. I tried opening combinations but didn't give me anything. Both combinatorial and algebraical proofs are welcomed. Any help or hint is so much appreciated!

Note: I want a more elementary way than using Laurent series method.

edited Nov 26 '23 at 18:42

RobPratt

45,619

asked Nov 26 '23 at 18:27

Mason Rashford

Hello there. Binomial coefficients scream for hypergeometric function of some sort. Don't know if that'll work, but nonetheless it is perhaps a way. – Vuk Stojiljkovic Nov 26 '23 at 18:30
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This is indeed related to Hypergeometric Identities (Although other combinatorial proofs may exist), refer $(2.6)$ Hypergeometric Identities – Miracle Invoker Nov 26 '23 at 18:39
There is an obvious conjecture for the product of $m$ terms with $(a_j){j=1}^m$ and $a{m+1}$ meaning $a_1$:
$\sum_{k} \prod_{j=1}^m \binom{a_j+a_{j+1}}{a_j+k}(-1)^k =\dfrac{(\sum_{j=1}^m a_j)!}{\prod_{j=1}^m a_j!} $. This is probably in Riordan or Knuth.
– marty cohen Nov 26 '23 at 18:41
This is called Dixon's Identity, BTW. – coiso Nov 26 '23 at 18:45
Compare https://math.stackexchange.com/q/338682/42969 and https://math.stackexchange.com/q/2119176/42969 – Martin R Nov 26 '23 at 18:57
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I have now added a much more elementary proof at the link How to show that $\sum_{k} (-1)^k{{a+b}\choose{a+k}}{{b+c}\choose{b+k}}{{c+a}\choose{c+k}} = \frac{(a+b+c)!}{a!b!c!}$ and vote to close this as a duplicate. Note that asking for more simplicity than that proof won't be possible : it only uses Chu-Vandermonde (which is a simple counting argument) and factorial manipulation, which is tedious but routine. – Sarvesh Ravichandran Iyer Nov 28 '23 at 06:11

Prove $\sum_k{{a+b \choose a+k}{b+c \choose b+k}{c+a \choose c+k}(-1)^k}=\dfrac{(a+b+c)!}{a!b!c!}$

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