Challenging interesting combinatorics problem

We can represent every combination of salaries with a quadruple $(a,b,c,d)$ , where each number is between 1 and 5; also, $a$ is the salary of the highest ranking employee, $b$ is the salary of the second-highest ranking employee, and so on.

Let $S$ be the set of all possible salaries, without any restrictions. Let $\begin{aligned} A &= \{(a,b,c,d) \in S : b < c < d\}, \\ B &= \{(a,b,c,d) \in S : a < c < d\}, \\ C &= \{(a,b,c,d) \in S : a < b < d\}, \\ D &= \{(a,b,c,d) \in S : a < b < c\}. \end{aligned}$ These sets represent the combinations we want to avoid; in other words, we want to count $|\overline{A} \cap \overline{B} \cap \overline{C} \cap \overline{D}|$ , which we can do by applying the Principle of Inclusion-Exclusion (PIE).

By PIE, $\begin{aligned} |\overline{A} \cap \overline{B} \cap \overline{C} \cap \overline{D}| &= |S| - |A| - |B| - |C| - |D| \\ &\quad + |A \cap B| + |A \cap C| + |A \cap D| + |B \cap C| + |B \cap D| + |C \cap D| \\ &\quad - |A \cap B \cap C| - |A \cap B \cap D| - |A \cap C \cap D| - |B \cap C \cap D| \\ &\quad + |A \cap B \cap C \cap D|. \end{aligned}$

It is easy to compute that $\begin{aligned} |S| &= 625, \\ |A| = |B| = |C| = |D| &= 50, \\ |A \cap B| = |B \cap C| = |C \cap D| &= 20, \\ |A \cap C| = |A \cap D| = |B \cap D| &= 5, \\ |A \cap B \cap C| = |A \cap B \cap D| = |A \cap C \cap D| = |B \cap C \cap D| &= 5, \\ |A \cap B \cap C \cap D| &= 5, \end{aligned}$ so $|\overline{A} \cap \overline{B} \cap \overline{C} \cap \overline{D}| = 625 - 4 \cdot 50 + (3 \cdot 20 + 3 \cdot 5) - 4 \cdot 5 + 5 = 485$ .