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Math Help - Combinatorial Identity, Pascal's Triangle, Hexagon Identity

  1. #1
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    Combinatorial Identity, Pascal's Triangle, Hexagon Identity

    Latex is pretty sweet! just learned how to use this, so for this question

    Prove the following combinatorial identity if {1}\leq k<n. This identity is known as the hexagon identity and relates terms in Pascal's Triangle.


    \left(\begin{array}{cc}n-1\\k-1\end{array}\right) \left(\begin{array}{cc}n\\k+1\end{array}\right) \left(\begin{array}{cc}n+1\\k\end{array}\right) = \left(\begin{array}{cc}n-1\\k\end{array}\right) \left(\begin{array}{cc}n\\k-1\end{array}\right) \left(\begin{array}{cc}n+1\\k+1\end{array}\right)


    where do i even begin to solve this? The Pascal's Triangle i kind of understand but the rest is news to me.
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    Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

    Quote Originally Posted by zhengcl86 View Post
    Latex is pretty sweet! just learned how to use this, so for this question

    Prove the following combinatorial identity if {1}\leq k<n. This identity is known as the hexagon identity and relates terms in Pascal's Triangle.


    \left(\begin{array}{cc}n-1\\k-1\end{array}\right) \left(\begin{array}{cc}n\\k+1\end{array}\right) \left(\begin{array}{cc}n+1\\k\end{array}\right) = \left(\begin{array}{cc}n-1\\k\end{array}\right) \left(\begin{array}{cc}n\\k-1\end{array}\right) \left(\begin{array}{cc}n+1\\k+1\end{array}\right)


    where do i even begin to solve this? The Pascal's Triangle i kind of understand but the rest is news to me.
    Use the identity

    \left(\begin{array}{cc}n\\k\end{array}\right) =\frac{n!}{(n-k)!k!}

    and this becomes trivially true by permutation of the denominators of the factors.
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  3. #3
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    Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

    Quote Originally Posted by Gusbob View Post
    Use the identity

    \left(\begin{array}{cc}n\\k\end{array}\right) =\frac{n!}{(n-k)!k!}

    and this becomes trivially true by permutation of the denominators of the factors.
    is that 1<k<n?
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  4. #4
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    Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

    Quote Originally Posted by zhengcl86 View Post
    Prove the following combinatorial identity if {1}\leq k<n. This identity is known as the hexagon identity and relates terms in Pascal's Triangle.


    \left(\begin{array}{cc}n-1\\k-1\end{array}\right) \left(\begin{array}{cc}n\\k+1\end{array}\right) \left(\begin{array}{cc}n+1\\k\end{array}\right) = \left(\begin{array}{cc}n-1\\k\end{array}\right) \left(\begin{array}{cc}n\\k-1\end{array}\right) \left(\begin{array}{cc}n+1\\k+1\end{array}\right)
    The LHS is
    \frac{(n-1)!}{(k-1)!(n-k)!}\cdot\frac{n!}{(k+1)!(n-k-1)!}\cdot\frac{(n+1)!}{(k!)(n-k+1)!}

    REARRANGE:
    \frac{(n-1)!}{(k)!(n-k-1)!}\cdot\frac{n!}{(k-1)!(n-k+1)!}\cdot\frac{(n+1)!}{(k+1)!(n-k)!}
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  5. #5
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    Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

    Quote Originally Posted by Plato View Post
    The LHS is
    \frac{(n-1)!}{(k-1)!(n-k)!}\cdot\frac{n!}{(k+1)!(n-k-1)!}\cdot\frac{(n+1)!}{(k!)(n-k+1)!}

    REARRANGE:
    \frac{(n-1)!}{(k)!(n-k-1)!}\cdot\frac{n!}{(k-1)!(n-k+1)!}\cdot\frac{(n+1)!}{(k+1)!(n-k)!}
    is the rearranged part the answer? i kinda have a similar answer but a couple faults on the 3rd spot where n+1 / (k+1) (n+1-k-1)

    somehow i simplified it down to (k-n)! (n-k-1)! = (n-k+1)! (n+1-k-1)!
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    Re: Combinatorial Identity, Pascal's Triangle, Hexagon Identity

    Quote Originally Posted by zhengcl86 View Post
    is the rearranged part the answer? i kinda have a similar answer but a couple faults on the 3rd spot where n+1 / (k+1) (n+1-k-1)

    somehow i simplified it down to (k-n)! (n-k-1)! = (n-k+1)! (n+1-k-1)!


    {\color{red}\frac{(n-1)!}{(k)!(n-k-1)!}}\cdot{\color{blue}\frac{n!}{(k-1)!(n-k+1)!}}\cdot\color{green}\frac{(n+1)!}{(k+1)!(n-k)!}

    {\color{red}\binom{n-1}{k}}\cdot{\color{blue}\binom{n}{k-1}}\cdot{\color{green}\binom{n+1}{k+1}}
    Thanks from Gusbob
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