There are two sections in a question paper each contains five questions. A…

Concept

When a fixed number of items must be selected and an upper limit caps how many may come from each group, the total count is the SUM over every admissible split of the products of independent choices: ways = Σ C(group₁, k₁) × C(group₂, k₂), taken over all splits (k₁, k₂) that satisfy the cap and total.

Application

Two sections of 5 questions each; answer 6 in all, with at most 4 from either section. Let x come from Section 1 and (6−x) from Section 2. The cap forces x ≤ 4 and 6−x ≤ 4, i.e. x ∈ {2, 3, 4}. Count each split:

1. (x = 2): C(5, 2) × C(5, 4) = 10 × 5 = 50

2. (x = 3): C(5, 3) × C(5, 3) = 10 × 10 = 100

3. (x = 4): C(5, 4) × C(5, 2) = 5 × 10 = 50

Adding the admissible splits: 50 + 100 + 50 = 200.

Cross-check

Without the cap, choosing any 6 of the 10 questions gives C(10, 6) = 210. The only forbidden splits are 5-from-one-section cases (x = 1 or x = 5), each C(5, 5) × C(5, 1) = 1 × 5 = 5, totalling 10. So 210 − 10 = 200 — the same result by an independent route.