We want to find the number of integer solutions of the equation
$$x_1 + x_2 + x_3 + x_4 = 30 \tag{1}$$
subject to the restrictions that $x_i \geq 2$, $x_i \neq 5$, for $1 \leq i \leq 4$.
As Severin Schraven pointed out in the comments, we can handle the first condition by defining $x_i = y_i + 2$ for $1 \leq i \leq 4$. Making these substitutions in equation 1 yields
\begin{align*}
y_1 + 2 + y_2 + 2 + y_3 + 2 + y_4 + 2 & = 30\\
y_1 + y_2 + y_3 + y_4 & = 22 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers. Since a particular solution corresponds to the placement of $4 - 1 = 3$ addition signs in a row of $22$ ones, the number of solutions of equation 2 in the nonnegative integers is
$$\binom{22 + 4 - 1}{4 - 1} = \binom{25}{3}$$
since we must choose which three of the $25$ positions required for $22$ ones and three addition signs will be filled with addition signs.
From these, we must subtract those solutions of equation 1 in which one or more of the $x_i$'s are equal to $5$ or, equivalently, those solutions of equation 2 in which one or more of the $y_i$'s is equal to $3$. Notice that at most three of the $x_i$'s may equal to $3$ since $4 \cdot 3 = 12 < 22$.
If $y_4 = 3$, then
\begin{align*}
y_1 + y_2 + y_3 + 3 & = 22\\
y_1 + y_2 + y_3 & = 19 \tag{3}
\end{align*}
Equation 3 is an equation in the nonnegative integers. Since a particular solution of equation 3 corresponds to the placement of $3 - 1 = 2$ addition signs in a row of $19$ ones, equation 3 has
$$\binom{19 + 3 - 1}{3 - 1} = \binom{21}{2}$$
solutions in the nonnegative integers. By symmetry, there are an equal number of solutions of equation 2 in which $y_1 = 3$ or $y_2 = 3$ or $y_3 = 3$ or $y_4 = 3$. Hence, the number of solutions in which one of the $y_i$'s is equal to $3$ is
$$\binom{4}{1}\binom{21}{2}$$
However, if we subtract this amount from the total, we will have subtracted too much since we will have subtracted each case in which two $y_i$'s are equal to three twice, once for each way we could have designated one of those $y_i$'s as being the one that is equal to $3$. We only want to subtract such cases once, so we must add them back.
If $y_3 = 3$ and $y_4 = 3$, then
\begin{align*}
y_1 + y_2 + 3 + 3 & = 22\\
y_1 + y_2 & = 16 \tag{4}
\end{align*}
Equation 4 is an equation in the nonnegative integers. Since a particular solution of equation 4 corresponds to the placement of $2 - 1 = 1$ addition sign in a row of $16$ ones, equation 4 has
$$\binom{16 + 2 - 1}{2 - 1} = \binom{17}{1}$$
solutions in the nonnegative integers. By symmetry, there are an equal number of solutions of equation 2 for each of the $\binom{4}{2}$ pairs $i, j$ such that $1 \leq i < j \leq 4$ in which $y_i = y_j = 3$. Hence, there are
$$\binom{4}{2}\binom{17}{1}$$
solutions of equation 2 in which two $y_i$'s are equal to $3$.
However, if we add these to our running total, we will have added too much. This is because we first subtracted those cases in which three $y_i$'s are equal to $3$ three times, once for each way we could designate one of the $y_i$'s as being the one that equals $3$, then subtracted them three times, once for each of the
$\binom{3}{2}$ ways we could designate a pair of the $y_i$'s as being the pair $(y_i, y_j)$ with $1 \leq i < j \leq 4$ in which $y_i = y_j = 3$. Hence, we have not subtracted those cases in which three of the $y_i$'s are equal to $3$.
If $y_2 = y_3 = y_4 = 3$, then
\begin{align*}
y_1 + 3 + 3 + 3 & = 22\\
y_1 & = 13 \tag{5}
\end{align*}
which is an equation in the nonnegative integers with one solution. By symmetry, there are an equal number of solutions for each of the $\binom{4}{3}$ ways of selecting triples $(i, j, k)$ with $1 \leq i < j < k \leq 4$ in which $y_i = y_j = y_k = 3$. Hence, there are
$$\binom{4}{3}\binom{13}{0}$$
solutions in which three of the $y_i$'s are equal to $3$.
By the Inclusion-Exclusion Principle, the number of admissible solutions of equation 1 is
$$\binom{25}{3} - \binom{4}{1}\binom{21}{2} + \binom{4}{2}\binom{17}{1} - \binom{4}{3}\binom{13}{0}$$
You have correctly reduced the problem to finding the number of solutions of the equation
$$y_1 + y_2 + y_3 + y_4 = 15 \tag{1}$$
subject to the restrictions $y_1 < 5, y_2 < 3, y_3 < 8, y_4 < 12$.
Let $A_1$ denote the set of outcomes in which $y_1 \geq 5$, $A_2$ denote the set of outcomes in which $y_2 \geq 3$, $A_3$ denote the set of outcomes in which $y_3 \geq 8$, and $A_4$ denote the set of outcomes in which $y_4 \geq 12$. By the Inclusion-Exclusion Principle, the number of outcomes in which none of the restrictions is violated is found by subtracting the number of solutions in which at least one of these restrictions is violated from the number of solutions of equation 1.
You correctly found that the number of solutions of equation 1 is
$$\binom{15 + 4 - 1}{4 - 1} = \binom{18}{3} = \binom{18}{15}$$
and that
\begin{align*}
|A_1| & = \binom{10 + 4 - 1}{4 - 1} = \binom{13}{3} = \binom{13}{10}\\
|A_2| & = \binom{12 + 4 - 1}{4 - 1} = \binom{15}{3} = \binom{15}{12}\\
|A_3| & = \binom{7 + 4 - 1}{4 - 1} = \binom{10}{3} = \binom{10}{7}\\
|A_4| & = \binom{3 + 4 - 1}{4 - 1} = \binom{6}{3}
\end{align*}
The reason you obtained a negative answer is that you have subtracted each case in which two restrictions are violated twice, once for each way you designated one of the restrictions as the restriction that is being violated. We only want to subtract such cases once, so we must add them to the total. In fact, by the Inclusion-Exclusion Principle, the number of solutions in which at least one condition is violated is
\begin{align*}
& |A_1 \cup A_2 \cup A_3 \cup A_4|\\
& \quad = |A_1| + |A_2| + |A_3| + |A_4|\\
& \qquad - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_1 \cap A_4| - |A_2 \cap A_3| - |A_2 \cap A_4| - |A_3 \cap A_4|\\
& \quad \qquad + |A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_4|\\
& \qquad \qquad - |A_1 \cap A_2 \cap A_3 \cap A_4|
\end{align*}
Notice that many of these terms are equal to zero. For instance, it is not possible for $y_1 \geq 5$ and $y_2 \geq 12$ since $5 + 12 > 15$.
Let's calculate $|A_1 \cap A_2|$. I will leave the calculations of the remaining terms to you.
$|A_1 \cap A_2|$: Then $y_1 \geq 5$ and $y_2 \geq 3$. Let $y_1' = y_1 - 5$ and $y_2' = y_2 - 3$. Then $y_1'$ and $y_2'$ are nonnegative integers. Substituting $y_1' + 4$ for $y_1$ and $y_2' + 3$ for $y_2$ in equation 1 yields
\begin{align*}
y_1' + 5 + y_2' + 3 + y_3 + y_4 & = 15\\
y_1' + y_2' + y_3 + y_4 & = 7 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{7 + 4 - 1}{4 - 1} = \binom{10}{3} = \binom{10}{7}$$
solutions.
Best Answer
It's a bit hard to follow your diagrams.
Sticking with what I think is your approach: There are $4$ gaps formed by your four choices, and the lengths of those gaps must sum to $11$. There are $\binom {10}3=120$ ways to populate those gaps with positive natural numbers. There are $15$ choices for the "first" selection. But of course any choice out of the $4$ could have been the "first" one so you are counting each selection $4$ times. Thus the desired result is $$\frac {\binom {10}3\times 15}{4}=450$$
Here's an example to illustrate the idea behind the calculation. Suppose the gap sequence was $\{3,3,3,2\}$. Those sum to $11$ so it's a valid choice. That alone does not give us a choice of four people though, you need to start somewhere. Say $1$ was the first selection. Then use that gap sequence to get the selection $(1,5,9,13)$. However you could have also obtained that selection by starting with $5$ and using the gap sequence $\{3,3,2,3\}$ or by starting with $9$ and using $\{3,2,3,3\}$ or by starting with $13$ and using $\{2,3,3,3\}$. Thus this procedure generates each selection exactly four times, so you must divide by $4$ to correct.