In a group of 3 girls and 4 boys, there are two boys \( B_1 \) and \( B_2 \). 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 \( B_1 \) and \( B_2 \) are not adjacent to each other, is:
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For problems involving arrangements with restrictions:
- Start by calculating the total number of arrangements without any restrictions.
- Then, subtract the cases where the restricted condition is violated (e.g., when \( B_1 \) and \( B_2 \) are adjacent).
- Use the principle of inclusion-exclusion if necessary.
Initially, the girls are treated as a single unit because they must stand together. This results in 5 entities to arrange: the block of girls and the 4 boys.The total permutations of these 5 entities is \( 5! \). However, it is required that \( B_1 \) and \( B_2 \) are not positioned next to each other.First, determine the total arrangements of the 5 entities:\[5! = 120.\]Next, calculate the arrangements where \( B_1 \) and \( B_2 \) are adjacent. By treating \( B_1 \) and \( B_2 \) as a single block, there are now 4 entities to arrange. The number of permutations for these 4 entities is \( 4! \), and within the \( B_1 B_2 \) block, \( B_1 \) and \( B_2 \) can be arranged in \( 2! \) ways.Therefore, the count of arrangements where \( B_1 \) and \( B_2 \) are adjacent is:\[4! \times 2! = 24 \times 2 = 48.\]The count of arrangements where \( B_1 \) and \( B_2 \) are not adjacent is:\[120 - 48 = 72.\]Consequently, the final answer is \( 72 \).
