Cominatorics (sticks and plus, generating function) If I should distribute 25 identical cookies to 10 children each children should get at least 1 cookie and maxium of 4 cookies. The problem should be solved using both inclusion-exclusion and generating function.
My solution with in-ex (Wrong): 
$$x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} +  x_{8} + x_{9} + x_{10} = 25$$ $,1\le x_{i} \ge 4 $
$$x_{i}= 1 +z_{i} $$
 $$z_{1}+1 + z_{2}+1 + z_{3}+1 + z_{4}+1 + z_{5} +1+ z_{6}+1 + z_{7}+1  +z_{8} +1+ z_{9}+1 + z_{10}+1 = 25$$
$$z_{1}  + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7}  +z_{8} + z_{9} + z_{10} = 15$$
${sticks+plus} \choose sticks  $$=$${24} \choose 15$. That is all cominations when everyone gets at least one.
The forbidden combinations, $z_{i} \ge 4 $,:$$z_{i}=4+u_{i}$$ $$4+u_{1}  + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7}  +z_{8} + z_{9} + z_{10} = 15$$
$$u_{1}  + z_{2} + z_{3} + z_{4} + z_{5} + z_{6} + z_{7}  +z_{8} + z_{9} + z_{10} = 11$$
${sticks+plus} \choose sticks  $$=$${20} \choose 11$, and this can be done in ten ways. So my answer is: ${24} \choose 15$$-10$${20} \choose 11$
According to the book this is wrong.
When doing it with GF I get: $X^{10}\frac{(1-x^{4})^{10}}{(1-x)^{10}}$. And then I don'tknow what to do. How to now which coefficient to find?

You have $[ x^{10} ]\cdot\frac{(1-x^4)^{10}}{(1-x)^{10}}$, where $[ x^{k} ]$ denote the coefficient of operator.
$\frac{1}{(1-x)^{10}}$ can be written as $\sum_{j=0}^\infty \binom{-10}{j}(-x)^j$ (Binomial series).
$$[ x^{10} ]\cdot(1-x^4)^{10}\cdot \sum_{j=0}^\infty \binom{-10}{j}(-x)^j$$
Now we use that $\binom{-r}{s}=\binom{r+s-1}{r-1}\cdot (-1)^s$
$$[ x^{10} ]\cdot(1-x^4)^{10}\cdot \sum_{j=0}^\infty \binom{9+j}{9}\cdot (x)^j$$
The coefficient of operator $[ x^{10} ] $can be combined with   $x^k, 0\leq k\leq 10$ as $[ x^{10} ]\cdot x^k=[ x^{10-k} ]$
$$\left([ x^{10} ]\cdot \binom{10}{0} -[ x^{6} ]\cdot \binom{10}{1}+[ x^{2} ]\cdot \binom{10}{2}-\ldots \right)\cdot \sum_{j=0}^\infty \binom{9+j}{9}\cdot (x)^j$$
Finally we choose the corresponding coefficients
$${19 \choose 9 }-10\cdot  {15 \choose 9} +45\cdot {11 \choose 9}=44,803$$
