Number of integer solutions in [0,10] How many solutions does the following equation have?
$$k+l+m+n=30,$$
where $k,l,m,n\in\mathbb{Z},0\leq k,l,m,n\leq 10.$
Edit: this is an exercise in the inclusion-exclusion principle.
 A: This is the coefficient of $x^{30}$ in $(1 + x + ... + x^{10})^4$.  (To see this, think about what happens when you choose a term from each factor.)  But this is $\frac{(1 - x^{11})^4}{(1 - x)^4}$.  Now,
$$\frac{1}{(1 - x)^4} = \sum_{n \ge 0} {n+3 \choose 3} x^n$$
(which you'll recognize as the generating function giving the answer to the problem without the range restriction.)  On the other hand, $(1 - x^{11})^4 = 1 - {4 \choose 1} x^{11} + {4 \choose 2} x^{22} - {4 \choose 3} x^{33} + x^{44}$.  Of these terms only the first three contribute to the coefficient of $x^{30}$, making the final answer
$${33 \choose 3} - {4 \choose 1} {22 \choose 3} + {4 \choose 2} {11 \choose 3} = 286.$$
This is equivalent to the following inclusion-exclusion argument: start with the set of all solutions with no restriction.  Remove all the solutions where one of the four numbers is greater than $10$.  Add back in all the solutions where two of the four numbers are greater than $10$.
A: A geometric way of seeing the answer is this: this is the number lattice points in the cube $[0,10]^4$ that intersects the plane $k+l+m+n = 30$. Note that the corners $(0,10,10,10)$ are points on this plane. Since the weights on $k,l,m,n$ are equal, the solutions are in bijection (it would be clear once you draw a picture) with the lattice points in the $k,l,m$ plane that is inside the tetrahedron formed by the vertices $(0,10,10)$, $(10,0,10)$, $(10,10,0)$ and $(0,0,0)$, the total number of which would be the 11th tetrahedral number which is 286. 
