An organization awarded 48 medals in event A ' ' , 25 in event B '...

Let $A, B$ and $C$ denote the set of men who received medals in even $A$, event $B$ and event $C$ respectively.

Then, $n\left(A\right) = 48, n\left(B\right) = 25, n\left(C\right) = 18$ , $n\left(A \cup B \cup C\right) = 60$ and $n\left(A \cap B \cap C\right) = 5$,

Now we know that, $n\left(A \cup B \cup C\right) = n\left(A\right) + n\left(B\right) + n\left(C\right) - n\left(A \cap B\right) - n\left(A \cap C\right) - n\left(B \cap C\right) + n\left(A \cap B \cap C\right)$

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

$⇒ n ( A \cap B ) + n ( A \cap C ) + n ( A \cap C ) = 48 + 25 + 18 + 5 - 60 = 36$

Therefore, the number of people who received medals in exactly two of the three sports will be $36 - 3\left(A \cap B \cap C\right) = 36 - 15 = 21$.