Counting number of solutions with restrictions
If you want $y\leq z$, use the substitution $z = y + z'$, and you're solving $x + 5y + 8(y+z') = n$, which is equivalent to non-negative integer solutions for $x + 13y + 8 z' = n$.
If you want $x\leq y \leq z$, a similar method will work. You could also do this for $2y \leq z$.
Of course, with more restrictions, this can get harder to clearly do. It could be difficult to solve subject to $x \leq y, z \leq 2y, x \geq \pi z$ (assuming any solutions exist).