(a) Case I When one question is selected from Section A, then there are 4 ways of selecting question from Section B i.e it can be selected either 1 question or 2 questions or 3 questions or 4 questions. The number of ways when one question from Section A is selected =3C1(4C1+4C2+4C3+4C4) =3(4+6+4+1) Similarly, the number of ways when two questions are selected from Section A =3C3(4C1+4C2+4C3+4C4) =3(4+6+4+1)=45 The number of ways when three questions are selected from Section A =3C3(4C1+4C2+4C3+4C4) =1(4+6+4+1)=15 ∴ The required number of ways =45+45+15=105