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Math Help - Binomial coefficient halp!

  1. #1
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    Binomial coefficient halp!

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

    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 x^{13} in the expansion of \left( {1 + x + x^2  + x^3  + x^4 } \right)^4 .
    We can get x^{13} from: \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: <br />
x^{12}  \cdot \left( {1 + x + ... + x^4 } \right)^4  = \frac{{x^{12}  \cdot \left( {1 - x^5 } \right)^4 }}<br />
{{\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)}}<br />
{{\left( {1 - x} \right)^4 }}<br />

    <br />
=<br />
 \frac{{x^{12} }}<br />
{{\left( {1 - x} \right)^4 }} - \frac{{4 \cdot x^{17} }}<br />
{{\left( {1 - x} \right)^4 }} + \frac{{6 \cdot x^{22} }}<br />
{{\left( {1 - x} \right)^4 }} - \frac{{4 \cdot x^{27} }}<br />
{{\left( {1 - x} \right)^4 }} + \frac{{x^{32} }}<br />
{{\left( {1 - x} \right)^4 }}<br />

    Now remember that: <br />
\frac{1}<br />
{{\left( {1 - x} \right)^4 }} = \sum\limits_{r = 0}^\infty  {\left( {\begin{array}{*{20}c}<br />
   {3 + r}  \\<br />
   r  \\<br /> <br />
 \end{array} } \right) \cdot x^r } <br />
( |x|<1)

    So we get; <br />
\sum\limits_{k = 12}^\infty  {\left( {\begin{array}{*{20}c}<br />
   {k - 9}  \\<br />
   {k - 12}  \\<br /> <br />
 \end{array} } \right)x^k }  - \sum\limits_{k = 17}^\infty  {4\left( {\begin{array}{*{20}c}<br />
   {k - 14}  \\<br />
   {k - 17}  \\<br /> <br />
 \end{array} } \right)x^k }  \pm ... + \sum\limits_{k = 32}^\infty  {\left( {\begin{array}{*{20}c}<br />
   {k - 29}  \\<br />
   {k - 32}  \\<br /> <br />
 \end{array} } \right)x^k } <br /> <br />

    For our coefficient is enough to consider: <br />
\sum\limits_{k = 12}^\infty  {\left( {\begin{array}{*{20}c}<br />
   {k - 9}  \\<br />
   {k - 12}  \\<br /> <br />
 \end{array} } \right)x^k }  - \sum\limits_{k = 17}^\infty  {4\left( {\begin{array}{*{20}c}<br />
   {k - 14}  \\<br />
   {k - 17}  \\<br /> <br />
 \end{array} } \right)x^k }  + \sum\limits_{k = 22}^\infty  {6\left( {\begin{array}{*{20}c}<br />
   {k - 19}  \\<br />
   {k - 22}  \\<br /> <br />
 \end{array} } \right)x^k } <br /> <br />

    And thus the coefficient is: <br />
\left( {\begin{array}{*{20}c}<br />
   {25 - 9}  \\<br />
   {25 - 12}  \\<br /> <br />
 \end{array} } \right) - 4 \cdot \left( {\begin{array}{*{20}c}<br />
   {25 - 14}  \\<br />
   {25 - 17}  \\<br /> <br />
 \end{array} } \right) + 6 \cdot \left( {\begin{array}{*{20}c}<br />
   {25 - 19}  \\<br />
   {25 - 22}  \\<br /> <br />
 \end{array} } \right)<br />
=<br />
\left( {\begin{array}{*{20}c}<br />
   {16}  \\<br />
   {13}  \\<br /> <br />
 \end{array} } \right) - 4 \cdot \left( {\begin{array}{*{20}c}<br />
   {11}  \\<br />
   8  \\<br /> <br />
 \end{array} } \right) + 6 \cdot \left( {\begin{array}{*{20}c}<br />
   6  \\<br />
   3  \\<br /> <br />
 \end{array} } \right)=20<br />

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

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