1. ## 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.

2. 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.

3. 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

4. thanks guys!

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