# The Hexagonal Property of Pascal's Triangle

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.

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

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, intereseted 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?

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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In symbols, the identity is

$$\left({n-1\atop m-1}\right)\left({n\atop m+1}\right)\left({n+1\atop m}\right) = \left({n\atop m-1}\right)\left({n-1\atop m}\right)\left({n+1\atop m+1}\right).$$

The usual combinatorial interpretation of a binomial coefficient $\left({n-1\atop m-1}\right)$ is that it counts subsets of size $m$ from a set of size $n$. Multiplication is usually interpreted as mutually exclusive choice ($f(n)g(n)$ counts the process of picking $f(n)$ configurations, then picking (independently) $g(n)$ items.

Putting this together, the LHS counts subsets of size $m-1$ from a set of size $n-1$, then subsets of size $m$ from an (independent) set of size $n+1$, then (again independently) subsets of size $m+1$ from a set of size $n$. This corresponds one-to-one with the RHS because the things counted by the LHS can be counted in a different way by the RHS: For the RHS distinguish an element of the $n$ set and one of the $n+1$ set. What's left over for those two sets can be chosen by $\left({n-1\atop (m+1)-1}\right)$ and $\left({(n+1)-1\atop m-1}\right)$ respectively, and then the two distinguished elements can be included to be (possibly) chosen in the $n-1$ set to account for $\left({(n-1) +2 \atop (m-1)+2}\right)$.

To be clearer about the combinatorial interpretation, there are three sets, of size $n-1$, $n$, and $n+1$, from which you choose subsets of size $m-1$, $m+1$, and $m$, respectively. Another way to count this situation is to, take 1 item each out of the $n$ and $n+1$ sets, and add them to the $n-1$ set. So now you're counting out of sets of size $n+1$, $n-1$, and $n$, from which you choose subsets of size $m+1$, $m$, and $m-1$, respectively.

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@milcak: I added a paragraph stating more clearly the combinatorial part. I am not using the arithmetic of factorials in my proof, simply using the combinatorial interpretation of binomial coefficients as counting subsets and counting one situation in two different ways. – Mitch Feb 7 '11 at 18:11
@milcak: Check out math.stackexchange.com/questions/15505/combinatorial-identity for a similarly styled combinatorial proof. – Mitch Feb 7 '11 at 19:02
@milcak This proof is completely combinatorial: it combinatorially interprets the two sides of the equality you want to prove and then combinatorially establishes the equality. I don't see what more you want. – Alex B. Feb 21 '11 at 5:20
@milcak: can you give an example of a proof of an identity involving just binomial coefficients that would be sufficient to be called 'combinatorial' by you? You may want to explain to what degree your example is 'pure' or avoids the use of 'how pascal's triangle is constructed'. (before dismissing my example, you should confirm that it does not use Pascal's identity) – Mitch Feb 23 '11 at 3:20