На третьем курсе студенты сдавали четыре экзамена....

Solving the combinatorics problem

The user is asking for the minimum percentage of students who could have passed all four exams, based on the pass rates for each. To approach this using inclusion-exclusion, I need to consider four sets with sizes 80%, 85%, 90%, and 98%. Essentially, I'll minimize the overlap between them, ensuring we stay within the bounds given by the individual percentages. The goal is to determine the smallest intersection between these sets, which will give us the minimum percentage who passed all exams.

Calculating minimal intersection

To minimize the number of students passing all exams, we treat the problem as finding the smallest overlap between four sets (representing the different exams). The general idea is to subtract the total of all percentages from 300, which gives a lower bound for the intersection — in this case, 53%. I’m now testing whether this result is achievable by allocating the remaining students to meet the pass requirements without exceeding the total count of students.

Thought for 37 seconds