Determine the number of integer solutions to $x_1 + x_2 + x_3 + x_4 = 19$, where $−5 \le x_i \le 10$ for all $1 \le i \le 4$ What I have so far:
Goal: Using the inclusion exclusion I want to find
$|\overline A_{1}\cap \overline A_{2} \cap \overline A_{3} \cap \overline A_{4}|  = |U| - S_{1} + S_{2} - S_{3} + S_{4}$
$S_{k} = \sum |\overline A_{i1}\cap \overline A_{i1} \cap ... \overline A_{ik}|$
I have incremented the values of i by 5 so that the range can start from zero like this:
$x_{1}+x_{2}+ x_{3} + x_{4} = 24$ with $0\leq x_{i} \leq 15$
For $|U|$ I have used to "stars and bars technique":
$|U| = \binom{r+n-1}{r} = \binom{24+4-1}{3} $
The Answer
... I am studying for a test (this is a practice question) and my professor has provided a solution that says: 
$\binom{42}{39} - \binom{4}{1} \binom{26}{23}+\binom{4}{2}\binom{10}{7}$
So I don't think I am on the right track if the universal set $|U| = \binom{42}{39}$. Any tips would be great thanks in advance.


I have incremented the values of i by 5 so that the range can start from zero like this: $
x_1+x_2+x_3+x_4=24$  with 
  $0≤x_i≤15$ 

Are you sure this is correct? You might need to check that inequality and the equation preceding it. 
