Inclusion-Exclusion How many integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 32$ with $0 \leq x_i \leq 10$ for $i = 1, 2, 3, 4$?
How many integer solutions are there to the inequality
$$
y_1 + y_2 + y_3 + y_4 < 184
$$
with $y_1 > 0$, $0 < y_2 \leq 10$, $0 \leq y_3 \leq 17$, and $0 \leq y_4 < 19$?
Here are two problems. I can see this combination $(32+4-1,4-1)$ but I am having trouble finding out what to subtract from that.
Also, the second problem is looking for a more complicated solution. I understand that $y_1+y_2+y_3+y_4+y_5 = 182$ and $(182,4)$ for the combination but what do it have to subtract from that?
Thank you.

Generating functions seem to give a cleaner looking solution for the second question.
The generating function for number of ways to sum to $k$ is
$$
\overbrace{\ \ \frac{x\vphantom{x^1}}{1-x}\ \ }^{y_1\gt0}\overbrace{\frac{x-x^{11}}{1-x}}^{0\lt y_2\le10}\overbrace{\frac{1-x^{18}}{1-x}}^{0\le y_3\le17}\overbrace{\frac{1-x^{19}}{1-x}}^{0\le y_4\lt19}
$$
which is
$$
\begin{align}
&\frac{x^2}{(1-x)^4}\left(1-x^{10}-x^{18}-x^{19}+x^{28}+x^{29}+x^{37}-x^{47}\right)\\
&=\left(x^2-x^{12}-x^{20}-x^{21}+x^{30}+x^{31}+x^{39}-x^{49}\right)\sum_{k=0}^\infty\binom{k+3}{k}x^k\\
&=\sum_{k=0}^\infty\left[{\scriptsize\binom{k+1}{k-2}-\binom{k-9}{k-12}-\binom{k-17}{k-20}-\binom{k-18}{k-21}+\binom{k-27}{k-30}+\binom{k-28}{k-31}+\binom{k-36}{k-39}-\binom{k-46}{k-49}}\right]x^k
\end{align}
$$
and the sum of the coefficients for $k\lt184$ is
$$
\begin{align}
&\sum_{k=0}^{183}\left[{\scriptsize\binom{k+1}{k-2}-\binom{k-9}{k-12}-\binom{k-17}{k-20}-\binom{k-18}{k-21}+\binom{k-27}{k-30}+\binom{k-28}{k-31}+\binom{k-36}{k-39}-\binom{k-46}{k-49}}\right]\\
&=\scriptsize\binom{185}{4}-\binom{175}{4}-\binom{167}{4}-\binom{166}{4}+\binom{157}{4}+\binom{156}{4}+\binom{148}{4}-\binom{138}{4}\\[6pt]
&=547200
\end{align}
$$
