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Math Help - Binomial summation proof

  1. #1
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    Binomial summation proof

    Show that:

    \sum_{k=m}^n \binom{k}{m}\binom{n}{k}p^k(1-p)^{n-k}=\binom{n}{m}p^m

    You may be able to use this result:

    \binom{k}{m}\binom{n}{k}=\binom{n}{m}\binom{n-m}{k-m}
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  2. #2
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    Quote Originally Posted by acevipa View Post
    Show that:

    \sum_{k=m}^n \binom{k}{m}\binom{n}{k}p^k(1-p)^{n-k}=\binom{n}{m}p^m

    You may be able to use this result:

    \binom{k}{m}\binom{n}{k}=\binom{n}{m}\binom{n-m}{k-m}
    Start with the Binomial Theorem:

    (x+y)^n = \sum_k \binom{n}{k}x^k y^{n-k}

    Differentiate m times with respect to x:

    (n-m+1) \cdots (n-1)(n) (x+y)^{n-m} = \sum_k \binom{n}{k} (k-m+1) \cdots (k-1)(k) x^{k-m}y^{n-k}

    Divide by m!:

    \binom{n}{m} (x+y)^{n-m} = \sum_k \binom{n}{k} \binom{k}{m} x^{k-m}y^{n-k}

    Now let x = p and y = 1-p:

    \binom{n}{m} (1)^{n-m} = \sum_k \binom{n}{k} \binom{k}{m} p^{k-m}(1-p)^{n-k}

    Multiply by p^m:

    \binom{n}{m} p^{m} = \sum_k \binom{n}{k} \binom{k}{m}  p^k (1-p)^{n-k}
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