Solution: Using the principle of inclusion-exclusion for three sets \( M \), \( P \), and \( C \), we have:
\[ |M \cup P \cup C| = |M| + |P| + |C| - |M \cap P| - |P \cap C| - |M \cap C| + |M \cap P \cap C| \]
Given:
Since \(|M \cup P \cup C| = 40\), substitute the values and solve for \(|M \cap P \cap C|\):
\[ 40 = 20 + 25 + 16 - 11 - 15 - 10 + x \]
\[ x = 10 \]
Thus, the maximum number of students who passed in all three subjects is 10.
Let \( S = \{p_1, p_2, \dots, p_{10}\} \) be the set of the first ten prime numbers. Let \( A = S \cup P \), where \( P \) is the set of all possible products of distinct elements of \( S \). Then the number of all ordered pairs \( (x, y) \), where \( x \in S \), \( y \in A \), and \( x \) divides \( y \), is _________.