Re: Divisibility property of binomial coefficients (Hoggatt triangles)

From: Martin Bundgaard (myinitials_at_adelic.org)
Date: 08/21/04 

Date: 21 Aug 2004 09:45:01 -0400

tchow@lsa.umich.edu wrote:

> Is it true, and if so how does one prove, that
>
> (n choose m) * (n choose m+1) * ... * (n choose m+k)
>
> is divisible by
>
> (n choose 0) * (n choose 1) * ... * (n choose k) ?
>

Yes, you have "factorwise" divisibility:

   (n choose m+k) k!(n-k)! (n-k)!
  ---------------- = --------------- = ----------- |m-k|!
    (n choose k) (m+k)!(n-m-k)! m!(n-m-k)!

  = (n-k choose m) |m-k|!

where we use (m+k)! = k!m!/|m-k|!.

--Martin