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In a group of 3 girls and 4 boys, there are two boys B₁ and B₂. The number of ways, in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but B₁ and B₂ are not adjacent to each other, is:
(1) 144
(2) 72
(3) 96
(4) 120

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Detailed Solution

Total arrangements where all girls are together and all boys are together: 2! × 3! × 4!

Arrangements where B₁ and B₂ are also adjacent: 2! × 3! × (3! × 2!)

Required arrangements = 2!(3! × 4!) - 2!(3! × 3! × 2!)

= 2 × 6 × 24 - 2 × 6 × 6 × 2

= 288 - 144 = 144

Answer: (1) 144

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