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A row of 7 seats is to be filled by 4 adults and 3 children so that no 2 children are seated next to each other. The 4 adults are distinct and the 3 children are distinct. In how many different ways can the 7 people be seated?
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Correct answer
C
Seat the 4 distinct adults first: 4! = 24 orders. The 4 adults create 5 gaps (one before each adult, one between each adjacent pair, and one after the last) where children may go. To keep children non-adjacent, place each child in a separate gap: choose 3 of the 5 gaps and order the 3 distinct children, P(5,3) = 5 × 4 × 3 = 60. Total = 24 × 60 = 1440.
Verification by an independent split of the gap step: choose which 3 of the 5 gaps hold a child, C(5,3) = 10; order the 3 distinct children among those gaps, 3! = 6; order the 4 adults, 4! = 24; product 10 × 6 × 24 = 1440, which matches.
Choice (A) 144 = 4! × 3! ignores the gap placement, (B) 720 drops the restriction, and (E) 5040 = 7! ignores it entirely.
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