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?

I find it a little clearer not to skip the first step, which is writing the generating function. 
you just write down a series $A(x)$ of formal variables $x^k$, 
where k denotes how many cookies each child becomes. 
As specified k must be between 1 and 4. 
Thus $$ A(x) = x+x^2+x^3+x^4 $$
As you have 10 children, the generating functions becomes, 
$$B(x) = \mathrm{A(x)}^{10} = \mathrm{(x+x^2+x^3+x^4)}^{10} = \frac{(1-x^4)^{10}}{ (1-x)^{10}} $$
Then extracting coefficients using convolution. 
