Imagine a class of 50 students who faced two examinations. In the first examination, 26 students achieved an 'A', while in the second examination, 21 of the students succeeded in earning an 'A'. Remarkably, 17 students did not manage to secure an 'A' in either of these two exams. The intriguing question arises: how many students excelled by earning an 'A' in both examinations?

Understanding the Problem

The inclusion-exclusion principle offers a sophisticated method to solve this type of logical and mathematical challenge. This principle is often used in set theory, combinatorics, and various fields such as computer science and economics.

Defining Sets

Let's define the following sets:

A?: The set of students who got an 'A' in the first examination. A?: The set of students who got an 'A' in the second examination.

From the problem, we understand that:

|A?| 26, meaning 26 students got an 'A' in the first examination. |A?| 21, indicating 21 students got an 'A' in the second examination. 17 students did not get an 'A' in either examination.

Note: The total number of students in the class is 50.

Calculating Students with 'A' in at Least One Exam

The number of students who got an 'A' in at least one of the two examinations can be found using the total number of students and the number who did not get an 'A' in either examination:

|A? ∪ A?| 50 - 17 33

Applying the Inclusion-Exclusion Principle

The principle of inclusion-exclusion states:

|A? ∪ A?| |A?| |A?| - |A? ∩ A?|

Substituting the known values:

33 26 21 - |A? ∩ A?|

Solving for |A? ∩ A?|:

|A? ∩ A?| 26 21 - 33 14

Conclusion

Therefore, the number of students who got an 'A' in both examinations is 14.

Additional Considerations

It is important to note that the principle of inclusion-exclusion provides a precise method when the sets are well-defined and their interactions are accounted for. The answer derived from this principle is 14, based on the given conditions for this particular problem.

For a greater class size where the principle may yield different results, let's consider two additional scenarios:

If the total number of students was 67, then all 21 students who received an 'A' in the second examination would have also received an 'A' in the first examination, because the number of students with an 'A' in at least one exam (33) would be insufficient to accommodate 21 new students who did not previously have an 'A'. If the total number of students was 88, then no student would have achieved an 'A' in both examinations because the total number of students with an 'A' in at least one exam (33) would be significantly fewer than the number needed for 21 students to have an 'A' in both exams.

Thus, the exact number of students who achieved an 'A' in both examinations, using the inclusion-exclusion principle, is 14, based on the given conditions in the problem statement.