An organization awarded 48 medals in event 'A', 25 in event 'B' and 18 in event 'C'. If these medals went to total 60 men and only five men got medals in all the three events, then how many received medals in exactly two of three events?

Using the inclusion-exclusion principle:

$$|A \cup B \cup C| = |A| + |B| + |C| - |A \cap B| - |B \cap C| - |A \cap C| + |A \cap B \cap C|$$

We are given that

$$|A| = 48$$, $$|B| = 25$$, $$|C| = 18$$

$$|A \cup B \cup C| = 60$$ (total men)

$$|A \cap B \cap C| = 5$$ (men with medals in all three events)

Substituting:

$$60 = 48 + 25 + 18 - (|A \cap B| + |B \cap C| + |A \cap C|) + 5$$

$$60 = 96 - (|A \cap B| + |B \cap C| + |A \cap C|)$$

$$|A \cap B| + |B \cap C| + |A \cap C| = 36$$

The number of men who received medals in exactly two events:

$$= (|A \cap B| + |B \cap C| + |A \cap C|) - 3|A \cap B \cap C|$$

$$= 36 - 3(5) = 36 - 15 = 21$$