How many solutions are there for $x + y + z = 12$? Where 
$$1 ≤ x ≤ 5 $$
$$y \ge 0$$
$$0 ≤ z ≤ 7$$
Without any restrictions I got the answer to be 91, then got amount of combinations with x greater than 5 where I got 28 and z greater than 7 which got me 15. I also attempted to get the combinations where x is less than 1 where I got 13. I subtracted all of these from the original 91 which got me 35. I was wondering how to account for the overlaps or if there was an easier way to combine all these steps, or if I'm just doing it wrong, thank you. 
 A: To elaborate, slightly, on the comment:  
With the restrictions as they stand there are $40$ solutions, not $35$.  To see that, note that there are $5$ possible values for $x$, and $8$ possible values for $z$ (trusting that all three were meant to be integers of course).  Choosing any of these pairs determines the value of $y$ and it is easy to see that all the $y$ so obtained satisfy the stated restriction.  Hence there are $5^*8=40$ solutions.
Note:  if the restriction on $z$ is modified to read $1≤z≤7$ then the same argument would yield $5^*7=35$ solutions, which is consistent with the stated answer.  Perhaps this is what was intended.
A: In this case, lulu's answer provides a perfectly effective solution, since the solution space essentially is a rectangular section of the $x+y+z = 12$ plane.
More generally, one way to do this is by stars and bars, first accounting for the minimum values of the three variables, and secondly accounting for the maximum values (plus one), and taking the difference between those to obtain the final count.
The example will clarify the approach: First account for the minimum values.  Let $x' = x+1$.  Then $x', y, z \geq 0$ and $x'+y+z = 11$.  By stars and bars, there are
$$
\binom{13}{2} = 78
$$
ways for this to happen.  However, this includes some ways in which $x > 5$, or where $z > 7$.  These must be subtracted out, individually, so we let $x'' = x+6$ and $x''+y+z = 6$; by stars and bars, this count is
$$
\binom{8}{2} = 28
$$
We let $z' = z+8$ and $x'+y+z' = 3$; the stars-and-bars count is
$$
\binom{5}{2} = 10
$$
In general, by inclusion-exclusion, we would now have to consider and add back in the cases where both $x > 5$ and $z > 7$; however, this cannot be the case with $y \geq 0$ and $x+y+z = 12$, so the effect here is zero (and lulu's much simpler approach thus works perfectly).
Therefore, the final count (assuming $z \geq 0$) is $78-28-10+0 = 40$.  If $z > 0$ instead, then the count is
$$
\binom{12}{2}-\binom{7}{2}-\binom{5}{2}+0 = 35
$$
