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.
 A: Note you can use the same method for the first question: if $x+y+z=8$ and $x,y,z>0 $, you can set $x=1+x', y=1+y', z=1+z'$, where $x',y',z'$ are nonnegative and $x'+y'+z'=5$. 
Therefore the number of positive solutions is 
$\;\dbinom{5+3-1}{3-1}=\dbinom 72=21.$
More generally, it is easy to prove  that the number of positive integer solutions of the equation
$$x_1+x_2+\dots +x_r=n \quad (n\ge r)$$
is equal to $\;\dbinom{n-1}{r-1}$.
A: For the second problem, write $x=-3+x'$ and so on. You have $x'+y'+z'=17$ and $x',\dots$ are nonnegative, a case you know how to solve.
You can also solve the first problem this way; now you would set $x=1+x'$, etc. 
A: 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 $z=Z-3$.
A: 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.
