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Binomial identities

  1. Apr 17, 2009 #1
    hello,

    i am supposed to show that

    Sigma of k = 0 to m, (n, k) (n - k, m - k) = 2^m (n, m)

    So I have after expanding:

    (n, k) = n!/(n-k)!k! and (n-k, m-k) = (n-k)!/(m-k)!(n-m)!
    so together the (n-k)! cancels out and I have
    n!/k!(m-k)!(n-m)!
    and that is
    n!/m!(n-m)! which is
    (n, m)

    so then I can take (n, m) out of the sigma and then I would have
    (n, m)*Sigma from k = 0 to m 1, but then how do I get the 2^m?

    Thanks.
     
  2. jcsd
  3. Apr 17, 2009 #2

    Dick

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    Science Advisor
    Homework Helper

    You seem to think that 1/(k!*(m-k)!) is 1/m!. It's not. Try some numerical examples. On the other hand it's easy to find the sum of that quantity over k. You probably proved sum over k of (m,k)=2^m, didn't you?
     
  4. Apr 17, 2009 #3
    Oh right. Thanks!!

    Yes, it's equal to (1 + 1) ^ m , so that is 2^m.
     
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