Arrangements With Together Condition
The number of ways to arrange 5 boys and 3 girls in a row so that all 3 girls always stand together as a single block equals which value?
Select the correct option:
Solution
4320
When a subset of objects must stay together, they are treated as one combined block, then permuted internally, a reliable JEE Advanced grouping method. Tie the 3 girls into a single block, so there are now 5 boys plus 1 girl-block, totalling 6 units to arrange in a row in 6! = 720 ways. Within the block, the 3 girls can be arranged among themselves in 3! = 6 ways. By the multiplication principle, the total is 720 × 6 = 4320. Option 720 counts only the block arrangements without internal girl orderings. Option 2880 uses an incorrect internal factor. Option 40320 = 8! ignores the togetherness condition entirely. Hence there are 4320 arrangements. Plausibility check: the count must be smaller than the unrestricted 8! = 40320, and 4320 is exactly 8!/ (something), specifically the fraction where the girls are bound, consistent with treating three of eight items as a single movable unit. The handshake count is the prototypical example of a triangular number arising from pair selection, and inverting it produces a quadratic whose positive root gives the group size. This same C(n,2) structure underlies counting edges of a complete graph, diagonals of polygons, and chords of a circle, making it one of the most transferable identities in combinatorics.
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About This Question
- Subject
- mathematics
- Chapter
- permutations and combinations
- Topic
- arrangements with together condition
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
4320
When a subset of objects must stay together, they are treated as one combined block, then permuted internally, a reliable JEE Advanced grouping method. Tie the 3 girls into a single block, so there are now 5 boys plus 1 girl-block, totalling 6 units to arrange in a row in 6! = 720 ways. Within the block, the 3 girls can be arranged among themselves in 3! = 6 ways. By the multiplication principle, the total is 720 × 6 = 4320. Option 720 counts only the block arrangements without internal girl orderings. Option 2880 uses an incorrect internal factor. Option 40320 = 8! ignores the togetherness condition entirely. Hence there are 4320 arrangements. Plausibility check: the count must be smaller than the unrestricted 8! = 40320, and 4320 is exactly 8!/ (something), specifically the fraction where the girls are bound, consistent with treating three of eight items as a single movable unit. The handshake count is the prototypical example of a triangular number arising from pair selection, and inverting it produces a quadratic whose positive root gives the group size. This same C(n,2) structure underlies counting edges of a complete graph, diagonals of polygons, and chords of a circle, making it one of the most transferable identities in combinatorics.
This medium difficulty mathematics question is from the chapter permutations and combinations, covering the topic of arrangements with together condition. It appeared in the 2025 exam.
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