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Math Help - Binomial coefficient

  1. #1
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    Binomial coefficient

    I want to proof

    \sum_{i=0}^{k}\left(\begin{array}{cc}m\\i\end{arra  y}\right)\left(\begin{array}{cc}{n}\\{k-i}\end{array}\right)=\left(\begin{array}{cc}{m+n}\  \k\end{array}\right)

    Can anyone help?
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  2. #2
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    Quote Originally Posted by bram kierkels View Post
    \sum_{i=0}^{k}\left(\begin{array}{cc}m\\i\end{arra  y}\right)\left(\begin{array}{cc}{n}\\{k-i}\end{array}\right)=\left(\begin{array}{cc}{m+n}\  \k\end{array}\right)
    Clearly 0\le k\le m+n.
    \binom{m+n}{k} is the number of ways to select k items from a set of m+n items.
    Let  A = \left\{ {1,2, \cdots ,m + n} \right\},\;B = \left\{ {1,2, \cdots ,k} \right\}\,\& \,C = A\backslash B.
    Can you see the sum as selecting i items from B and k-i items from C?
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  3. #3
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    Quote Originally Posted by Plato View Post
    Clearly 0\le k\le m+n.
    \binom{m+n}{k} is the number of ways to select k items from a set of m+n items.
    Let  A = \left\{ {1,2, \cdots ,m + n} \right\},\;B = \left\{ {1,2, \cdots ,k} \right\}\,\& \,C = A\backslash B.
    Can you see the sum as selecting i items from B and k-i items from C?
    I see what you mean. But shouldn't B have m elements? So that B = \left\{ {1,2, \cdots ,m}\right\} and C=A\backslash B becomes \left\{ {m+1,m+2, \cdots ,m+n} \right\}
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  4. #4
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    Quote Originally Posted by bram kierkels View Post
    So that B = \left\{ {1,2, \cdots ,m}\right\}
    No that will not work,
    What happens if k>m?
    That gives an error in the summation.
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