Label the pools: 4 affirmative-only (A), 3 negative-only (N), 2 either-side (E). The panel needs 2 on affirmative and 2 on negative, and an E member counts as a different panel depending on which side you assign. Split on how many E members are used.
No E member: 2 from A and 2 from N gives C(4,2) × C(3,2) = 6 × 3 = 18.
Exactly one E member: pick which one (2 ways), then place it. On affirmative needs 1 more from A and 2 from N: C(4,1) × C(3,2) = 12. On negative needs 2 from A and 1 from N: C(4,2) × C(3,1) = 18. Each E member contributes 12 + 18 = 30, and there are 2 of them: 60.
Both E members: both on affirmative needs 2 from N = 3; both on negative needs 2 from A = 6; one on each side gives 2 assignments times 1 more from A (4) × 1 more from N (3) = 24. That totals 33.
Grand total: 18 + 60 + 33 = 111. So the answer is D.
E (150) is the naive product C(6,2) × C(5,2), which double-counts panels that reuse an either-side member on both sides. B (60) keeps only the exactly-one-E case. C (99) counts unordered groups of 4 people and forgets that groups containing both E members plus one of each fixed type can be assigned two ways. A (18) forgets the either-side members exist at all. The choices are bare counts, so confirming any of them requires rebuilding the three cases.