We have, part A has 7 questions, Part B has 5 questions and part
C has 3 questions.
Case I 4 questions from part
A, then 3 questions from part
B and
CNumber of ways
=7C4×[5C3+5C2×3C1+5C1×3C2]=[10+30+15]=35×55=1925Case II 3 questions from Part
A, then 4 questions from part
B and
C.
∴ Number of ways
=7C3×[5C3×3C1+5C2×3C2]=35×[30+30]=35×60=2100Case III 2 questions from part
A, then 5 question from part
A and
BNumber of ways
=7C2[5C3×3C2]=21×10×3=630∴ Required number of ways
=1925+2100+630=4655
