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Thread: Binomial coefficient halp!

  1. #1
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    Binomial coefficient halp!

    can someone help me find the coefficient of $\displaystyle x^{25} $ in :

    $\displaystyle x^{12}(1+x+x^2+x^3+x^4)^4 $

    I cant seem to find how to switch it up, I know the answer but just now how to get to it.
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  2. #2
    MHF Contributor

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    You want the coefficient of $\displaystyle x^{13}$ in the expansion of $\displaystyle \left( {1 + x + x^2 + x^3 + x^4 } \right)^4 $.
    We can get $\displaystyle x^{13}$ from: $\displaystyle \left( x \right)\left( {x^4 } \right)\left( {x^4 } \right)\left( {x^4 } \right)\,,\,\left( {x^3 } \right)\left( {x^3 } \right)\left( {x^3 } \right)\left( {x^4 } \right)\,,\mbox{or} \left( {x^2 } \right)\left( {x^3 } \right)\left( {x^4 } \right)\left( {x^4 } \right)$
    Find out how many ways to rearrange each of those strings and add it up.
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  3. #3
    Super Member PaulRS's Avatar
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    Note that: $\displaystyle
    x^{12} \cdot \left( {1 + x + ... + x^4 } \right)^4 = \frac{{x^{12} \cdot \left( {1 - x^5 } \right)^4 }}
    {{\left( {1 - x} \right)^4 }} = \frac{{x^{12} \cdot \left( {1 - 4 \cdot x^5 + 6 \cdot x^{10} - 4 \cdot x^{15} + x^{20} } \right)}}
    {{\left( {1 - x} \right)^4 }}
    $

    $\displaystyle
    =
    \frac{{x^{12} }}
    {{\left( {1 - x} \right)^4 }} - \frac{{4 \cdot x^{17} }}
    {{\left( {1 - x} \right)^4 }} + \frac{{6 \cdot x^{22} }}
    {{\left( {1 - x} \right)^4 }} - \frac{{4 \cdot x^{27} }}
    {{\left( {1 - x} \right)^4 }} + \frac{{x^{32} }}
    {{\left( {1 - x} \right)^4 }}
    $

    Now remember that: $\displaystyle
    \frac{1}
    {{\left( {1 - x} \right)^4 }} = \sum\limits_{r = 0}^\infty {\left( {\begin{array}{*{20}c}
    {3 + r} \\
    r \\

    \end{array} } \right) \cdot x^r }
    $ ($\displaystyle |x|<1$)

    So we get; $\displaystyle
    \sum\limits_{k = 12}^\infty {\left( {\begin{array}{*{20}c}
    {k - 9} \\
    {k - 12} \\

    \end{array} } \right)x^k } - \sum\limits_{k = 17}^\infty {4\left( {\begin{array}{*{20}c}
    {k - 14} \\
    {k - 17} \\

    \end{array} } \right)x^k } \pm ... + \sum\limits_{k = 32}^\infty {\left( {\begin{array}{*{20}c}
    {k - 29} \\
    {k - 32} \\

    \end{array} } \right)x^k }

    $

    For our coefficient is enough to consider: $\displaystyle
    \sum\limits_{k = 12}^\infty {\left( {\begin{array}{*{20}c}
    {k - 9} \\
    {k - 12} \\

    \end{array} } \right)x^k } - \sum\limits_{k = 17}^\infty {4\left( {\begin{array}{*{20}c}
    {k - 14} \\
    {k - 17} \\

    \end{array} } \right)x^k } + \sum\limits_{k = 22}^\infty {6\left( {\begin{array}{*{20}c}
    {k - 19} \\
    {k - 22} \\

    \end{array} } \right)x^k }

    $

    And thus the coefficient is: $\displaystyle
    \left( {\begin{array}{*{20}c}
    {25 - 9} \\
    {25 - 12} \\

    \end{array} } \right) - 4 \cdot \left( {\begin{array}{*{20}c}
    {25 - 14} \\
    {25 - 17} \\

    \end{array} } \right) + 6 \cdot \left( {\begin{array}{*{20}c}
    {25 - 19} \\
    {25 - 22} \\

    \end{array} } \right)
    $$\displaystyle =
    \left( {\begin{array}{*{20}c}
    {16} \\
    {13} \\

    \end{array} } \right) - 4 \cdot \left( {\begin{array}{*{20}c}
    {11} \\
    8 \\

    \end{array} } \right) + 6 \cdot \left( {\begin{array}{*{20}c}
    6 \\
    3 \\

    \end{array} } \right)=20
    $

    A longer way, but still the same result
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  4. #4
    Member
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    Jan 2008
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    thanks guys!

    Didnt think it would be that difficult to be honest with you guys.
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