<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] Your answer to the question of how many solutions the equation $$x + y + z = 8 \tag{1} $$ has in the positive integers is incorrect.  As a sanity check, observe that there must be fewer solutions to the equation in the positive integers than there are in the nonnegative integers since we are not allowed to substitute $0$ for any of the variables.

How many integer solutions are there to the equation $x + y + z = 8$ when $x, y, z > 0$?

If $x, y, z$ are positive integers, then $x' = x - 1$, $y' = y - 1$, and $z' = z - 1$ are nonnegative integers.  Substituting $x' + 1$ for $x$, $y' + 1$ for $y$, and $z' + 1$ for $z$ in equation 1 yields
\begin{align*}
x' + 1 + y' + 1 + z' + 1 & = 8\\
x' + y' + z' & = 5 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{5 + 3 - 1}{3 - 1} = \binom{7}{2}$$
solutions.  Notice that there are fewer solutions to equation 1 in the positive integers than the nonnegative integers, as we would expect.

How many integer solutions are there to the equation $x + y + z = 8$ when $x, y, z \geq -3$?

If $x, y, z \geq -3$, then $x' = x + 3$, $y' = y + 3$, and $z' = z + 3$ are nonnegative integers.  Substitute $x' + 3$ for $x$, $y' + 3$ for $y$, and $z' + 3$ for $z$ in equation 1, then proceed as above.
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