Key Concepts and Formulas

• Princ of Inclusion-Exclusion (PIE) for three sets: $|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |C \cap A|) + |A \cap B \cap C|$ This formula counts the total number of distinct elements in the union of three sets.
• Elements in Exactly Two Sets: The number of elements belonging to exactly two of the three sets (A, B, C) can be derived from the PIE. If $S_2 = |A \cap B| + |B \cap C| + |C \cap A|$, then the number of elements in exactly two sets is $S_2 - 3 \times |A \cap B \cap C|$.

Step-by-Step Solution

Step 1: Define Sets and List Given Information Let A, B, and C be the sets of men who received medals in event 'A', event 'B', and event 'C', respectively. We are given:

• $|A| = 48$ (Number of medals in event A)
• $|B| = 25$ (Number of medals in event B)
• $|C| = 18$ (Number of medals in event C)
• $|A \cup B \cup C| = 60$ (Total number of unique medal recipients)
• $|A \cap B \cap C| = 5$ (Number of men who received medals in all three events)

We need to find the number of men who received medals in exactly two of the three events.

Step 2: Apply the Principle of Inclusion-Exclusion to find the sum of pairwise intersections. The Principle of Inclusion-Exclusion states: $|A \cup B \cup C| = |A| + |B| + |C| - (|A \cap B| + |B \cap C| + |C \cap A|) + |A \cap B \cap C|$ Let $S_2 = |A \cap B| + |B \cap C| + |C \cap A|$. This term represents the sum of men who received medals in at least two events. Rearranging the PIE formula to solve for $S_2$: $S_2 = |A| + |B| + |C| + |A \cap B \cap C| - |A \cup B \cup C|$ Substitute the given values: $S_2 = 48 + 25 + 18 + 5 - 60$ $S_2 = 96 - 60$ $S_2 = 36$ This means the sum of men in (A and B), (B and C), and (C and A) is 36.

Step 3: Calculate the number of men who received medals in exactly two events. The sum $S_2$ counts men who received medals in exactly two events once, and men who received medals in all three events three times (once for each pair: $A \cap B$, $B \cap C$, $C \cap A$). To find the number of men who received medals in exactly two events, we must subtract the overcounted men from $S_2$. Men who received medals in all three events ($|A \cap B \cap C|$) were counted 3 times in $S_2$, but they should not be counted at all in the "exactly two" category. Therefore, we subtract $3 \times |A \cap B \cap C|$ from $S_2$. Number of men in exactly two events = $S_2 - 3 \times |A \cap B \cap C|$ Substitute the calculated and given values: Number of men in exactly two events = $36 - 3 \times 5$ Number of men in exactly two events = $36 - 15$ Number of men in exactly two events = $21$

Common Mistakes & Tips

• Distinguishing "At Least Two" from "Exactly Two": The sum of pairwise intersections ($|A \cap B| + |B \cap C| + |C \cap A|$) counts individuals in all three events multiple times. This sum is not the answer itself.
• Correcting for Triple Intersection: When calculating "exactly two," remember that each person in $|A \cap B \cap C|$ is included in all three pairwise intersections. Thus, you must subtract $3 \times |A \cap B \cap C|$ from the sum of pairwise intersections.
• Understanding PIE Components: Clearly identify what each term in the PIE formula represents to avoid misinterpretation, especially $|A \cup B \cup C|$ (total unique individuals) and $|A \cap B \cap C|$ (individuals in all categories).

Summary

By applying the Principle of Inclusion-Exclusion, we first calculated the sum of men who received medals in at least two events. We then adjusted this sum to account for the men who received medals in all three events, as they were overcounted in the pairwise intersections. This adjustment involved subtracting three times the number of men who received medals in all three events, leading to the final count of men who received medals in exactly two events.

The final answer is $\boxed{21}$ which corresponds to option (C).