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 . This identity is known as the hexagon identity and relates terms in Pascal's Triangle.

=

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

. This identity is known as the hexagon identity and relates terms in Pascal's Triangle.

=

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

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

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**

The LHS is

REARRANGE:

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

Quote:

Originally Posted by

**Plato** The LHS is

REARRANGE:

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)!