1. 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

No clue yet?

3. Re: Combination identity

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
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: $\displaystyle {2^n} = \sum\limits_{k = 0}^n {\binom{n}{k}}$.

But there is no such for what you ask.

4. Re: Combination identity

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

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

But there is no such for what you ask.
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).

Im aware that such result is not useful practically.
But it could help proving other identities or equations.

5. 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?

Thank you for your help

6. Re: Combination identity

Im very sorry.
Someone pointed out to my mistake.
Here is my new formula (corrected)