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?

$\mathrm{n}\left(\mathrm{A}\right)=48, \mathrm{n}\left(\mathrm{B}\right)=25, \mathrm{n}\left(\mathrm{C}\right)=18$

$\mathrm{Total} \mathrm{men}=60=\mathrm{n}(\mathrm{A}\cup \mathrm{B}\cup \mathrm{C})$

$\mathrm{n}(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C})=5$

$$\begin{gathered} \mathrm{n}(\mathrm{A}\cup \mathrm{B}\cup \mathrm{C})=\mathrm{n}\left(\mathrm{A}\right)+\mathrm{n}\left(\mathrm{B}\right)+\mathrm{n}\left(\mathrm{C}\right)-\mathrm{n}(\mathrm{A}\cap \mathrm{B})-\mathrm{n}(\mathrm{A}\cap \mathrm{C})-\mathrm{n}(\mathrm{B}\cap \mathrm{C}) \\ +\mathrm{n}(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C}) \end{gathered}$$

$$\begin{gathered} 60=48+25+18-\left[\mathrm{n}(\mathrm{A}\cap \mathrm{B})+\mathrm{n}(\mathrm{B}\cap \mathrm{C})+\mathrm{n}(\mathrm{A}\cap \mathrm{C})\right]+5 \\ \Rightarrow \mathrm{n}(\mathrm{A}\cap \mathrm{B})+\mathrm{n}(\mathrm{A}\cap \mathrm{C})+\mathrm{n}(\mathrm{B}\cap \mathrm{C})=36 \end{gathered}$$
Exactly two medals received in 3 events 
$$\begin{gathered} =\mathrm{n}(\mathrm{A}\cap \mathrm{B})+\mathrm{n}(\mathrm{B}\cap \mathrm{C})+\mathrm{n}(\mathrm{A}\cap \mathrm{C})-3\mathrm{n}(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C}) \\ =36-3\left(5\right) = 21 \end{gathered}$$