In a class of $$30$$  pupils, $$12$$ take needle work, $$16$$  take physics and $$18$$ take history. If all the $$30$$  students take at least one subjects take and no one takes all three then the number of pupils taking $$2$$ subjects is :

Solution :
$$\eqalign{ & {\text{Given }}n\left( N \right) = 12,\,n\left( P \right) = 16,\,n\left( H \right) = 18,\,n\left( {N \cup P \cup H} \right) = 30 \cr & {\text{From, }}n\left( {N \cup P \cup H} \right) = n\left( N \right) + n\left( P \right) + n\left( H \right) - n\left( {N \cap P} \right) - n\left( {P \cap H} \right) - n\left( {N \cap H} \right) + n\left( {N \cap P \cap H} \right) \cr & \therefore \,n\left( {N \cap P} \right) + n\left( {P \cap H} \right) + n\left( {N \cap H} \right) = 16 \cr} $$
Now, number of pupils taking two subjects
$$\eqalign{ & = n\left( {N \cap P} \right) + n\left( {P \cap H} \right) + n\left( {N \cap H} \right) - 3n\left( {N \cap P \cap H} \right) \cr & = 16 - 0 \cr & = 16 \cr} $$