An organization awarded $48$ medals in event ${}^{\text{'}}A^{\text{'}}$, $25$ in event ${}^{\text{'}}B^{\text{'}}$ and $18$ in event ${}^{\text{'}}C^{\text{'}}$. 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?

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

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

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

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