I am trying to solve the following question. I have worked through a similar question but the question I have worked through the range was all positive, however this one has a negative side of the range so I am not quite sure how to solve it. I will include the question I have worked through as reference below. Here is the question I need help solving, thanks! I need to solve it through the principle of inclusion exclusion
How many integer solutions of $x_1+x_2+x_3+x_4=28$ are there with $-10 \leq x_i \leq 20$
Here is the similar problem I mentioned above,
How many integer solutions for the equation $x_1+x_2+x_3+x_4=28$ are there with $0 \leq x_i \leq 10$
For this solution I used the negation $N(a_1',a_2'a_3'a_4') = N -N(a_i)+N(a_i,a_j)-...$
I let $N=$ all possible solutions. I said it was like distrubting 28 balls in 4 bins so, $\binom{28+4=1}{28}$
the next would term would happen after I use $11$ balls, one more than the range of the question as listed above which is $10$, so we have $\binom{17+4-1}{17}$ and there are $\binom{4}{1}$ also, so $\binom{17+4-1}{17} \binom{4}{1}$
next after using 11 more balls we would have $\binom{6+4-1}{6}$ with $\binom{4}{2}$ so putting together we have $\binom{6+4-1}{6} \binom{4}{2}$
the term in the sequence is not possible since $6-11$ makes us run out of balls
so then the solution is, $\binom{28+4=1}{28} - \binom{17+4-1}{17} \binom{4}{1} + \binom{6+4-1}{6} \binom{4}{2}$
So then how can I use that similar process to solve my question