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?