Question
In how many ways can 5 boys and 3 girls be arranged in a row so that no two girls sit together?
Solution
ATQ, Arrange 5 boys first: 5! = 120 ways. This creates 6 possible gaps for girls: _ B _ B _ B _ B _ B _ → 6 gaps. Choose 3 of these 6 gaps for the 3 girls: C(6,3) = 20. Arrange 3 girls among themselves: 3! = 6. Total arrangements = 5! × C(6,3) × 3! = 120 × 20 × 6 = 14,400.
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