<s>[INST] How many integer solutions are there to $x+y+z=8$ 
How many integer solutions are there to $x+y+z=8$ When $x,y,z>0$? When $x,y,z\geq -3$?

So I know there is a formula for computing the number of nonnegative solutions
${8+3-1 \choose 3-1}={10\choose 2}$
So I then just subtracted cases where one or two integers are $0$.
If just $x=0$ then there are $6$ solutions where neither $y,z=0$.
So I multiplied this by $3$, then added the cases where two integers are $0$
$3\cdot 6+3=21$. So I get ${10 \choose 2}+21=66$
For the last problem where $x,y,z\geq -3$ I'm not sure how to deal with it.
 [/INST] Start by looking at the number of solutions of
\begin{eqnarray*}
X+Y+Z=17
\end{eqnarray*}
which has $\binom{17+3-1}{3-1}$ solutions.
This is the same as the number of solutions of
\begin{eqnarray*}
x+y+z=8
\end{eqnarray*}
where $x=X-3$,$y=Y-3$ and $z=Z-3$.
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