Combinatorial Identity, Pascal's Triangle, Hexagon Identity

Latex is pretty sweet! just learned how to use this, so for this question

Prove the following combinatorial identity if $\displaystyle {1}\leq k<n$. This identity is known as the hexagon identity and relates terms in Pascal's Triangle.

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

where do i even begin to solve this? The Pascal's Triangle i kind of understand but the rest is news to me.

Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

Quote:

Originally Posted by

**zhengcl86** Latex is pretty sweet! just learned how to use this, so for this question

Prove the following combinatorial identity if $\displaystyle {1}\leq k<n$. This identity is known as the hexagon identity and relates terms in Pascal's Triangle.

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

where do i even begin to solve this? The Pascal's Triangle i kind of understand but the rest is news to me.

Use the identity

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

and this becomes trivially true by permutation of the denominators of the factors.

Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

Quote:

Originally Posted by

**Gusbob** Use the identity

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

and this becomes trivially true by permutation of the denominators of the factors.

is that 1<k<n?

Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

Quote:

Originally Posted by

**zhengcl86** Prove the following combinatorial identity if $\displaystyle {1}\leq k<n$. This identity is known as the hexagon identity and relates terms in Pascal's Triangle.

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

The LHS is

$\displaystyle \frac{(n-1)!}{(k-1)!(n-k)!}\cdot\frac{n!}{(k+1)!(n-k-1)!}\cdot\frac{(n+1)!}{(k!)(n-k+1)!}$

REARRANGE:

$\displaystyle \frac{(n-1)!}{(k)!(n-k-1)!}\cdot\frac{n!}{(k-1)!(n-k+1)!}\cdot\frac{(n+1)!}{(k+1)!(n-k)!}$

Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

Quote:

Originally Posted by

**Plato** The LHS is

$\displaystyle \frac{(n-1)!}{(k-1)!(n-k)!}\cdot\frac{n!}{(k+1)!(n-k-1)!}\cdot\frac{(n+1)!}{(k!)(n-k+1)!}$

REARRANGE:

$\displaystyle \frac{(n-1)!}{(k)!(n-k-1)!}\cdot\frac{n!}{(k-1)!(n-k+1)!}\cdot\frac{(n+1)!}{(k+1)!(n-k)!}$

is the rearranged part the answer? i kinda have a similar answer but a couple faults on the 3rd spot where n+1 / (k+1) (n+1-k-1)

somehow i simplified it down to (k-n)! (n-k-1)! = (n-k+1)! (n+1-k-1)!

Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

Quote:

Originally Posted by

**zhengcl86** is the rearranged part the answer? i kinda have a similar answer but a couple faults on the 3rd spot where n+1 / (k+1) (n+1-k-1)

somehow i simplified it down to (k-n)! (n-k-1)! = (n-k+1)! (n+1-k-1)!

$\displaystyle {\color{red}\frac{(n-1)!}{(k)!(n-k-1)!}}\cdot{\color{blue}\frac{n!}{(k-1)!(n-k+1)!}}\cdot\color{green}\frac{(n+1)!}{(k+1)!(n-k)!}$

$\displaystyle {\color{red}\binom{n-1}{k}}\cdot{\color{blue}\binom{n}{k-1}}\cdot{\color{green}\binom{n+1}{k+1}}$