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Thread: Combinatorics

  1. #1
    Junior Member
    Joined
    Mar 2011
    Posts
    31

    Combinatorics

    Show that:

    $\displaystyle \[ \left( \begin{array}{cc}
    n \\
    k \\ \end{array} \right)\]$$\displaystyle \[ \left( \begin{array}{cc}
    n-k \\
    j \\ \end{array} \right)\]$=$\displaystyle \[ \left( \begin{array}{cc}
    n \\
    j \\ \end{array} \right)\]$$\displaystyle \[ \left( \begin{array}{cc}
    n-j \\
    k \\ \end{array} \right)\]$


    $\displaystyle j+k\leq n$

    need some help here, only thing that came to my mind was that

    $\displaystyle \[ \left( \begin{array}{cc}
    n \\
    k \\ \end{array} \right)\]$=$\displaystyle \[ \left( \begin{array}{cc}
    n \\
    n-k \\ \end{array} \right)\]$

    then what?

    Thanks!
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
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    Thanks
    848
    Hello, mechaniac!

    $\displaystyle \displaystyle \text{Show that: } {n\choose k}{n-k\choose j} \;=\; {n\choose j}{n-j\choose k} $

    Start with the left side:

    $\displaystyle \displaystyle {n\choose k}{n\!-\!k\choose j} \;=\;\frac{n!}{k!\,(n\!-\!k)!}\cdot\frac{(n\!-\!k)!}{j!(n\!-\!j\!-\!k)!} \;=\;\frac{n!}{j!\,k!\,(n\!-\!j\!-\!k)!} $


    $\displaystyle \text{Multiply by }\dfrac{(n\!-\!j)!}{(n\!-\!j)!}$

    . . $\displaystyle \displaystyle \frac{(n\!-\!j)!}{(n\!-\!j)!}\cdot\frac{n!}{j!\,k!\,(n\!-\!j\!-\!k)!} \;=\;\frac{n!\,(n\!-\!j)!}{j!\,k!\,(n\!-\!j)!(n\!-\!j\!-\!k)!} $


    . . $\displaystyle \displaystyle =\;\frac{n!}{j!(n\!-\!j)!} \cdot \frac{(n\!-\!j)!}{k!\,(n\!-\!j\!-\!k)!} \;=\;{n\choose j}{n\!-\!j\choose k}$

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  3. #3
    Junior Member
    Joined
    Mar 2011
    Posts
    31
    of course! thanks for the good explanation!
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