# hexagon identity

• May 31st 2008, 10:18 AM
robocop_911
hexagon identity
Can anyone prove the below given identity?
I am giving it a try I also want someone else to verify!

Suppose that k and n are integers with 1<k<=n. Prove the hexagon identity
which relates terms in Pascal's triangle that form a hexagon.

$\displaystyle \left(\begin{array}{cc}n-1\\k-1\end{array}\right) \left(\begin{array}{cc}n\\k+1\end{array}\right) \left(\begin{array}{cc}n+1\\k\end{array}\right) = \left(\begin{array}{cc}n-1\\k\end{array}\right) \left(\begin{array}{cc}n\\k-1\end{array}\right) \left(\begin{array}{cc}n+1\\k+1\end{array}\right)$
• May 31st 2008, 10:27 AM
topsquark
Quote:

Originally Posted by robocop_911
Can anyone prove the below given identity?
I am giving it a try I also want someone else to verify!

Suppose that k and n are integers with 1<k<=n. Prove the hexagon identity
which relates terms in Pascal's triangle that form a hexagon.

$\displaystyle \left(\begin{array}{cc}n-1\\k-1\end{array}\right) \left(\begin{array}{cc}n\\k+1\end{array}\right) \left(\begin{array}{cc}n+1\\k\end{array}\right) = \left(\begin{array}{cc}n-1\\k\end{array}\right) \left(\begin{array}{cc}n\\k-1\end{array}\right) \left(\begin{array}{cc}n+1\\k+1\end{array}\right)$

The easiest way is to just write out the terms for each side:
Left hand side:
$\displaystyle {{n - 1} \choose {k - 1}} {{n} \choose {k + 1}} {{n + 1} \choose {k}} = \frac{(n - 1)!}{(k - 1)!(n - k)!} \cdot \frac{n!}{(k + 1)!(n - k - 1)!} \cdot \frac{(n + 1)!}{k!(n + 1 - k)!}$

You do the right hand side.

By the way, check out how I did the coding. It's much easier.

-Dan
• May 31st 2008, 10:35 AM
robocop_911
Quote:

Originally Posted by topsquark
The easiest way is to just write out the terms for each side:
Left hand side:
$\displaystyle {{n - 1} \choose {k - 1}} {{n} \choose {k + 1}} {{n + 1} \choose {k}} = \frac{(n - 1)!}{(k - 1)!(n - k)!} \cdot \frac{n!}{(k + 1)!(n - k - 1)!} \cdot \frac{(n + 1)!}{k!(n + 1 - k)!}$

You do the right hand side.
-Dan

Can we "simplify" this identity a bit more? Or we just have to convert this identity into "factorial" form and we are done?
• May 31st 2008, 10:41 AM
topsquark
Quote:

Originally Posted by robocop_911
Can we "simplify" this identity a bit more? Or we just have to convert this identity into "factorial" form and we are done?

Why simplify? All you need to do is show that both sides are equal. When you write out the right hand side you will see that you have the same terms in the numerator and denominator.

-Dan