Find the number of integer solutions of $|x|+|y| \le 10$ 
Find the solutions of $|x|+|y| \le 10$, where $x$ and $y$ are integers.

My solution:
$$|x|+|y|+z=10$$
Now the solutions if there were no absolute values is:
$$\binom{13-1}{10}=\frac{11*12}{2}=66$$
now subtract that once that have $0$ then multiply the others by $4$ and  multiply that ones that only has $1$ zeroes by $2$.I get $221$ Am I right?

Pick's Theorem says that the area of a polygon whose vertices have integer coordinates is given by
$$A=I+{B\over2}-1$$
where $I$ is the number of Interior points with integer coordinates and $B$ is the number of Boundary points with integer coordinates.  For the given problem the polygon is a square with diagonals of length $20$, so $A=200$.  The bounday points satisfy $|x|+|y|=10$, so it's easy to see that $B=40$, hence $I=200-{40\over2}+1=181$.  The number we want is
$$I+B=181+40=221$$
