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Thread: Binomial coefficient

  1. #1
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    Binomial coefficient

    I want to proof

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

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