In an examination, 10 questions are to be answered, choosing at least 4 from…

Concept: When a fixed number of selections must be split between two independent groups subject to a minimum-per-group limit, first enumerate every valid split of the total between the two groups, then count each group's choices with the combination formula nCr = n! / (r!(n-r)!), and finally add the case-wise products by the addition principle, since the cases are mutually exclusive.

Application:

1. Part A and Part B each have 6 questions, and exactly 10 of these 12 must be answered, with at least 4 taken from each part.

2. The only splits (Part A count, Part B count) that add to 10 while keeping each part's count between 4 and 6 are (4, 6), (5, 5), and (6, 4).

3. For the split (4, 6): number of ways = C(6, 4) x C(6, 6) = 15 x 1 = 15.

4. For the split (5, 5): number of ways = C(6, 5) x C(6, 5) = 6 x 6 = 36.

5. For the split (6, 4): number of ways = C(6, 6) x C(6, 4) = 1 x 15 = 15.

6. Since these three splits cannot occur together, add them: 15 + 36 + 15 = 66.

Cross-check: swapping Part A and Part B turns the (4, 6) split into the (6, 4) split, so both must give the same count -- both equal 15, confirming no error there; re-adding 15 + 36 + 15 independently again gives 66, matching the total.