Given $$n\left( U \right) = 20,\,n\left( A \right) = 12,\,n\left( B \right) = 9,\,n\left( {A \cap B} \right) = 4,$$           where $$U$$ is the universal set, $$A$$ and $$B$$ are subsets of $$U$$, then $$n\left( {{{\left( {A \cup B} \right)}^C}} \right) = ?$$

Solution :
$$\eqalign{ & n\left( {A \cup B} \right) = n\left( A \right) + n\left( B \right) - n\left( {A \cap B} \right) \cr &= 12 + 9 - 4 \cr &= 17 \cr & {\text{Now, }}n\left( {{{\left( {A \cup B} \right)}^C}} \right) = n\left( U \right) - n\left( {A \cup B} \right) \cr &= 20 - 17 \cr &= 3 \cr} $$