An organization awarded 48 medals in event A, | vedclass

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(C) Let $|A|=48$,$|B|=25$,and $|C|=18$. Given the total number of men who received at least one medal is $|A \cup B \cup C|=60$. Given the number of m

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$21$

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(C) Let $|A|=48$,$|B|=25$,and $|C|=18$. Given the total number of men who received at least one medal is $|A \cup B \cup C|=60$. Given the number of men who received medals in all three events is $|A \cap B \cap C|=5$. Using the principle of inclusion-exclusion: $|A \cup B \cup C| = (|A| + |B| + |C|) - (|A \cap B| + |B \cap C| + |C \cap A|) + |A \cap B \cap C|$. Let $S_1 = |A| + |B| + |C| = 48 + 25 + 18 = 91$. Let $S_2 = |A \cap B| + |B \cap C| + |C \cap A|$. Then $60 = 91 - S_2 + 5$. $S_2 = 91 + 5 - 60 = 36$. The number of men who received medals in exactly two events is given by the formula: $N(	ext{exactly two}) = S_2 - 3|A \cap B \cap C|$. $N(	ext{exactly two}) = 36 - 3(5) = 36 - 15 = 21$.

An organization awarded $48$ medals in event $A$,$25$ in event $B$,and $18$ in event $C$. If these medals were awarded to a total of $60$ men and only $5$ men received medals in all three events,how many men received medals in exactly two of the three events?

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