Permutations and Combinations

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 C
Number of ways
‌=‌7C4×[‌5C3+‌5C2×‌3C1+‌5C1×‌3C2]
‌=‌

7×6×5

1×2×3

[10+30+15]
‌=35×55=1925
Case 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=2100
Case III 2 questions from part A, then 5 question from part A and B
Number of ways =‌7C2[‌5C3×‌3C2]
=21×10×3=630
∴ Required number of ways
=1925+2100+630=4655