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Math Help - Binomial Theorem and Sum Conversion

  1. #1
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    Binomial Theorem and Sum Conversion

    Hi,

    I have been asked to derive a simple closed form expression for the
    following sum:

    n
    ∑ (-1)^(m+i) *(nCi)*(iCm) = S n, m
    i = m

    [with m = 0,1,2,....,n ; n= 0,1,2,....] and C is the combinations symbol.


    I know that we are meant to be using the binomial theorem but I dont quite know where to go from there. Any help would be much appreciated.
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  2. #2
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    Quote Originally Posted by Number Cruncher 20 View Post
    Hi,

    I have been asked to derive a simple closed form expression for the
    following sum:

    n
    ∑ (-1)^(m+i) *(nCi)*(iCm) = S n, m
    i = m

    [with m = 0,1,2,....,n ; n= 0,1,2,....] and C is the combinations symbol.


    I know that we are meant to be using the binomial theorem but I dont quite know where to go from there. Any help would be much appreciated.
    Three hints:

    1. Prove that \sum_{i=0}^{n} (-1)^i {n \choose i} is equal to zero if n \neq 0 and is equal to 1 if n = 0.

    2. Prove that {n \choose k} {k \choose l} = {n \choose l} {n - l \choose k - l} for l \leq k.

    3. Deduce from 1. and 2. that your expression is equal to zero if m \neq n and is equal to 1 if n = m.
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  3. #3
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    Joined
    May 2008
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    Yes I see now, thanks a lot for the help!
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