# Combination identity

• Apr 11th 2013, 08:08 PM
Mouhaha
Combination identity
Hi everyone,
Im trying to find a way to express the combination formula C(n,k)=n!/(k!(n-k)!) as summation.
Is there any known identity filling that condition?

Thank you for any clue
• Apr 12th 2013, 06:07 PM
Mouhaha
Re: Combination identity
No clue yet?
• Apr 12th 2013, 06:25 PM
Plato
Re: Combination identity
Quote:

Originally Posted by Mouhaha
Hi everyone,
Im trying to find a way to express the combination formula C(n,k)=n!/(k!(n-k)!) as summation.
Is there any known identity filling that condition

Quote:

Originally Posted by Mouhaha
No clue yet?

There is "No clue yet?" because there is no connection whatsoever.

There are several standard identities such as: ${2^n} = \sum\limits_{k = 0}^n {\binom{n}{k}}$.

But there is no such for what you ask.
• Apr 13th 2013, 06:52 AM
Mouhaha
Re: Combination identity
Quote:

Originally Posted by Plato
There is "No clue yet?" because there is no connection whatsoever.

There are several standard identities such as: ${2^n} = \sum\limits_{k = 0}^n {\binom{n}{k}}$.

But there is no such for what you ask.

You expressed 2^n as sum of combinations. I know the expansion.
My goal is different.
I tried to find how to express C(n,k) as summation.

Here is what I mean.

If we call k the degree of C(n,k) I have found an identity such as :

C(n,k)= sum (a(i)*C(b(i),k-2)) i varying from b to d

I have found it while working on C(n,5).

I tried to google without success.

We could expand any C(n,k) by recurrence as sum of products.

C(n,k) .....> C(m,k-2)
C(m,k-2) .....> C(l,k-4)
and so on

I checked it for C(n,5).

Im aware that such result is not useful practically.
But it could help proving other identities or equations.
• Apr 13th 2013, 02:40 PM
Mouhaha
Re: Combination identity
Hi,

As I do not know how to use Latex here is the formula I discovered recently (look at the attachment).

What I want to know :
Is the formula correct?
How to prove it?
Is the formula known?
What can we do with such formula?