Hi everyone,

I`m 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

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- April 11th 2013, 08:08 PMMouhahaCombination identity
Hi everyone,

I`m 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 - April 12th 2013, 06:07 PMMouhahaRe: Combination identity
No clue yet?

- April 12th 2013, 06:25 PMPlatoRe: Combination identity
- April 13th 2013, 06:52 AMMouhahaRe: Combination identity
Thank you for your comment.

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).

I`m aware that such result is not useful practically.

But it could help proving other identities or equations. - April 13th 2013, 02:40 PMMouhahaRe: 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?

Thank you for your help - April 14th 2013, 04:39 PMMouhahaRe: Combination identity
I`m very sorry.

Someone pointed out to my mistake.

Here is my new formula (corrected)