Permutations Avoiding Adjacency
The number of ways to arrange 4 boys and 4 girls in a row so that no two girls are adjacent to each other equals which value?
Select the correct option:
Solution
2880
Arranging so that certain items are never adjacent uses the gap method: place the unrestricted items first, then slot the restricted items into the gaps, a powerful JEE Advanced technique. First arrange the 4 boys in a row in 4! = 24 ways. These boys create 5 gaps, including the two ends, into which girls may be placed: B_B_B_B. To ensure no two girls are adjacent, place the 4 girls into 4 of these 5 gaps, at most one girl per gap, in P(5,4) = 5·4·3·2 = 120 ways. By the multiplication principle, the total is 24 × 120 = 2880. Option 1152 uses an incorrect gap count. Option 5760 double counts. Option 40320 = 8! ignores the adjacency restriction. Hence there are 2880 arrangements. Plausibility check: the gap method guarantees separation because each girl occupies a distinct gap between or beside boys, and 24 × 120 = 2880 is well below the unrestricted 8! = 40320, consistent with the strong non-adjacency constraint. The gap method enforces non-adjacency by seating the unrestricted items first and dropping the restricted items into the gaps they create, guaranteeing separation by construction. It is the strategic mirror of the block method for togetherness, and the count of available gaps, one more than the number of seated items, is the detail that must be tracked precisely to avoid error.
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About This Question
- Subject
- mathematics
- Chapter
- permutations and combinations
- Topic
- permutations avoiding adjacency
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
2880
Arranging so that certain items are never adjacent uses the gap method: place the unrestricted items first, then slot the restricted items into the gaps, a powerful JEE Advanced technique. First arrange the 4 boys in a row in 4! = 24 ways. These boys create 5 gaps, including the two ends, into which girls may be placed: B_B_B_B. To ensure no two girls are adjacent, place the 4 girls into 4 of these 5 gaps, at most one girl per gap, in P(5,4) = 5·4·3·2 = 120 ways. By the multiplication principle, the total is 24 × 120 = 2880. Option 1152 uses an incorrect gap count. Option 5760 double counts. Option 40320 = 8! ignores the adjacency restriction. Hence there are 2880 arrangements. Plausibility check: the gap method guarantees separation because each girl occupies a distinct gap between or beside boys, and 24 × 120 = 2880 is well below the unrestricted 8! = 40320, consistent with the strong non-adjacency constraint. The gap method enforces non-adjacency by seating the unrestricted items first and dropping the restricted items into the gaps they create, guaranteeing separation by construction. It is the strategic mirror of the block method for togetherness, and the count of available gaps, one more than the number of seated items, is the detail that must be tracked precisely to avoid error.
This hard difficulty mathematics question is from the chapter permutations and combinations, covering the topic of permutations avoiding adjacency. It appeared in the 2025 exam.
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