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Yes that's true there is a nice symmetry under $r$ and $s$! Maybe my question should be "why do the triads behave so nicely?" Could you please give an insight into that? Or am I not thinking deeply enough about the anwsers already given to me??
@Theo No, I do not get turned off by such things - on these sites people do not know each others background, and so I understand why he posted that. However, his proof dissasambles the triads of the hexagons. I understand how you can see it is combinatorial, I just want something that uses the entire triad to count something, somehow.
The proof starts by expressing the property in a formula. I find that none combinatorial right away. Furthermore, to establish the equality he uses the same arguement 3 times, namely, a variation on an in-or-out arguement. I do believe there is a more intrinsic justification of this property. I'm looking for something such as one triple in the hexagon counts ..., the other ..., without actually coming back to each binomial independently.
