Combinatorics How many solutions for x+y+z+w=10 0<=x,y,z <= 6 2<=w<=5 Well, The question is in the Title.
It is part of my homework,
I had to give a solution with Generating Functions and I reached one, I'm just not sure if it is right and would appericiate any help solving the problem in different way so I can compare the final answer. 
Please do not solve it in Generating Functions, because If my answer was different then yours, I will try again.
The question again :
How many Integer solutions for $$x+y+z+w=10 $$while :
$$0 \leq x\leq y\leq z \leq 6$$ 
 $$2 \leq w \leq 5$$

Use generating functions. For example, $0 \le x \le 6$ gets represented by:
$$
1 + z + z^2 + \ldots + z^6
  = \frac{1 - z^7}{1 - z}
$$
Similarly, for $2 \le w \le 5$:
$$
z^2 + z^3 + z^4 + z^5
  = \frac{z^2 (1 - z^4)}{1 - z}
$$
The complete set is the product of the above, you want the coefficient of $z^{10}$. Use ellipsis for terms that don't affect the result:
\begin{align}
[z^{10}] \frac{(1 - z^7)^3 z^2 (1 - z^4)}{(1 - z)^4}
  &= [z^8] \frac{(1 - 3 z^7 + \ldots) (1 - z^4)}{(1 - z)^4} \\
  &= [z^8] (1 - z^4 - 3 z^7 + \ldots) (1 - z)^{-4} \\
  &= [z^8] (1 - z)^{-4} - [z^4] (1 - z)^{-4} - 3 [z] (1 - z)^{-4}
\end{align}
Use of the generalized binomial theorem finishes this off.
