1. ## 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.

2. 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.

3. Note that: $
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 }}
$

$
=
\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: $
\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 }
$
( $|x|<1$)

So we get; $
\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: $
\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: $
\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)
$
$=
\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

4. thanks guys!

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