Any hexagon in Pascal's triangle, whose vertices are 6 binomial coefficients surrounding any entry, has the property that:

the product of non-adjacent vertices is constant.

the greatest common divisor of non-adjacent vertices is constant.

Here is one such hexagon. As an example, we have that $4 \cdot 10 \cdot 15 = 6 \cdot 20 \cdot 5$, as well as $\gcd(4, 10, 15) = \gcd(6,20,5)$.

$$ 1 \\ 1 \qquad 1\\ 1\qquad 2\qquad 1\\ 1\qquad3\qquad3\qquad1\\ 1\qquad\mathbf{4}\qquad\mathbf{6}\qquad4\qquad1\\ 1\qquad\mathbf{5}\qquad10\qquad\mathbf{10}\qquad5\qquad1 \\ 1\qquad6\qquad\mathbf{15}\qquad\mathbf{20}\qquad15\qquad6\qquad1$$

There is a quick proof here (pdf). The original proof should be in *V. E. Hoggatt, Jr., & W. Hansell. "The Hidden Hexagon Squares." The Fibonacci Quarterly 9(1971):120, 133.* but I cannot access it.

I am, however, interested in a purely *combinatorial* proof. I do not know how to approach this at all: I cannot see what the non-adjacent vertices represent and/or I do not know how to remodel their meaning. Can anyone help?

I have asked this question on math.se, I have not yet received a satisfactory anwser. (The anwser provided there, is *somewhat* combinatorial in spirit, but maybe *one level down,* from what I am looking for. It begins with "In symbols, the identity is..." and then only uses the in-or-out arguement to finish up. It would be perfect if someone shared a problem/application of this too.)

**EDIT:** To specify my question more closely, what I am looking for is some natural bijection between the two sets of triads that create the hexagon.

Thanks.

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