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Thread: Combinatorics.

  1. July 20th 2010, 02:15 AM #16
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    Quote Originally Posted by ENRIQUESTEFANINI View Post
    sum_{k=1..n-1} C(n-1,k-1)C(n,k) = C(2n-1,n-1). So |T_4| = 35.
    When n=4, sum_{k=1..n-1} C(n-1,k-1)C(n,k) = C(3,0)C(4,1) + C(3,1)C(4,2) + C(3,2)C(4,3) = 4 + 18 + 12 = 34 \ne 35
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  2. July 20th 2010, 02:29 AM #17
    ENRIQUESTEFANINI
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    It's true. The "identity" you quoted in post #16 is not really an identity. Thanks and regards.
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