<s>[INST] Inclusion–Exclusion Identical Computers Problem 
Find the number of ways to distribute 19 identical computers to four schools, if School A must get at least three, School B must get
  at least two and at most five, School C get at most four, and School D gets the rest.

a) Solve using inclusion-exclusion
b) Solve using generating functions
I've been tackling this question for a couple of days and I am pretty confused where to even start. 
So I bet the answer is in the form $x_1 + x_2 + x_3 + x_4 = 19$,
where $x_1 \ge 3; 2 \le x_2 \le 5; x_3 \le 4$ and $x_4$ is whatever is left over.
And then I get confused. I've looked all over the internet, through textbooks and I'm not getting anywhere. Any help would be appreciated, thank you.
 [/INST] Addendum to the generating function part of the answer of @callculus. Assuming WA is not available, it's not too hard to calculate the coefficient by hand. In order to do so it's convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series. This way we can write e.g.
\begin{align*}
[x^j](1+x)^n=\binom{n}{j}
\end{align*}

We obtain
  \begin{align*}
[x^{19}]&\left(\sum_{k=3}^{17}x^k\right)\left(\sum_{k=2}^5x^k\right)\left(\sum_{k=0}^4x^k\right)\left(\sum_{k=0}^{14}x^k\right)\\
&=[x^{19}]\left(\sum_{k=0}^{14}x^{k+3}\right)\left(\sum_{k=0}^3x^{k+2}\right)\frac{1-x^5}{1-x}\cdot\frac{1-x^{15}}{1-x}\tag{1}\\
&=[x^{19}]x^3\frac{1-x^{15}}{1-x}\cdot x^2\frac{1-x^4}{1-x}\cdot\frac{1-x^5}{1-x}\cdot\frac{1-x^{15}}{1-x}\tag{2}\\
&=[x^{14}]\left(1-x^{15}\right)^2\left(1-x^4\right)\left(1-x^5\right)\sum_{n=0}^{\infty}\binom{-4}{n}(-x)^n\tag{3}\\
&=[x^{14}]\left(1-x^4\right)\left(1-x^5\right)\sum_{n=0}^{\infty}\binom{-4}{n}(-x)^n\tag{4}\\
&=[x^{14}](1-x^4-x^5+x^9)\sum_{n=0}^{\infty}\binom{n+3}{3}x^n\tag{5}\\
&=\left([x^{14}]-[x^{10}]-[x^9]+[x^5]\right)\sum_{n=0}^{\infty}\binom{n+3}{3}x^n\tag{6}\\
&=\binom{17}{3}-\binom{10}{3}-\binom{9}{3}+\binom{8}{3}\tag{7}\\
&=680-286-220+56\\
&=230
\end{align*}

Comment:


*

*In (1) we shift the index of $k$ to start from $0$ and apply the finite geometric series formula.

*In (2) we apply the finite geometric series formula again.

*In (3) we use the linearity of the coefficient of operator and apply the rule $$[x^{n-m}]A(x)=[x^n]x^mA(x)$$ We also use the binomial series expansion of $\frac{1}{(1-x)^4}$.

*In (4) we observe that we can replace the binom $(1-x^{15})$ with $1$, since multiplication with $x^{15}$ does nothing contribute to the coefficient of $x^{14}$.

*In (5) we use the binomial identity
\begin{align*}
  \binom{-p}{q}=\binom{p+q-1}{p-1}(-1)^q
  \end{align*}

*In (6) we apply the rule $[x^{n-m}]A(x)=[x^n]x^mA(x)$ again.

*In (7) we select the coefficients accordingly.
</s>