For the n-dimensional Torus, the k-th homology group (with integer coefficients) is isomorphic to the direct sum of $n \choose k$ copies of $\mathbb{Z}$. Poincaré duality thus gives us a somewhat indirect way to show that ${n \choose k} = {n \choose n-k}$.

I am looking for results (or a starting point in the literature) where less obvious combinatorial results are derived by Poincaré duality.

I think that this is roughly related to the following questions:

How can we realize different combinatorial objects as the dimension of a construction on vector spaces? Are the resulting algebras useful?

Combinatorial results without known combinatorial proofs

From what I have learned from the comments, the Dehn–Sommerville equations would be an example of a combinatoric result based on Poincaré duality.

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    $\begingroup$ Since this question doesn't have a unique correct answer, I'm going to flag it for community-wiki. $\endgroup$ – HJRW Jun 18 '15 at 8:31
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    $\begingroup$ Someone who knows about toric varieties surely has something to say here... (Poincaré duality holds for these with rational coefficients) $\endgroup$ – Dylan Wilson Jun 18 '15 at 14:20
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    $\begingroup$ If you're willing to extend this from "Poincar\'e duality" to "properties of Betti numbers", then you could include Hard Lefschetz in your toolset, and get Stanley's proof of the Upper Bound Theorem for the number of faces of a simplicial polytope. $\endgroup$ – Allen Knutson Jun 18 '15 at 14:32
  • $\begingroup$ I am willing to extend to any other duality concepts in (co-)homology. $\endgroup$ – Rolf Bardeli Jun 18 '15 at 14:49
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    $\begingroup$ Stanley's survey paper on unimodal sequences is a possible starting point. dedekind.mit.edu/~rstan/pubs/pubfiles/72.pdf However, I don't think that it gives any examples other than the ones already mentioned. In general, I think it would be very unusual for the symmetry of a sequence to be provable most easily by a topological duality theorem, rather than by a direct combinatorial argument. Even the Dehn--Sommerville equations have a relatively straightforward combinatorial proof. $\endgroup$ – Timothy Chow Jun 18 '15 at 16:13

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