Difficulty: Hard
Correct Answer: 7! x 6!
Explanation:
Introduction / Context:
This is a circular permutation problem with restrictions. We have 7 teachers and 6 students, all distinct individuals, to be seated around a round table. The key condition is that no two students may sit next to each other. Circular arrangements are counted up to rotation (that is, rotating everyone together is considered the same arrangement), and this changes the counting formula compared to linear arrangements. The “gap method” is the standard approach for problems where a subset of people cannot sit together.
Given Data / Assumptions:
- Distinct people: 7 teachers (T) and 6 students (S), total 13 persons.
- Seating is around a circular table, so rotations of the same relative order are not counted separately.
- No two students may sit adjacent to each other.
- All teachers and students are distinguishable from one another.
Concept / Approach:
To ensure that no two students are adjacent, we first arrange the teachers around the table and then place the students in the “gaps” between teachers. Using circular permutation logic, we fix the relative order of teachers first. Once teachers are seated, they create specific gaps around the table where students can be placed without violating the adjacency rule. Each gap can contain at most one student if we want to avoid two students sitting together.
Step-by-Step Solution:
Step 1: Arrange the 7 teachers around the circular table.Step 2: The number of distinct circular arrangements for 7 distinct teachers is (7 - 1)! = 6!.Step 3: Once teachers are seated, there are exactly 7 gaps between them (one between each pair of adjacent teachers, going around the circle).Step 4: To prevent students from sitting together, at most one student may occupy each gap.Step 5: We must place 6 distinct students into these 7 gaps, choosing 6 of the 7 gaps and ordering the students among those chosen gaps.Step 6: The number of ways to choose which 6 gaps will be occupied is 7C6 = 7.Step 7: Once the 6 gaps are chosen, the 6 students can be permuted among these 6 positions in 6! ways.Step 8: Total number of valid circular arrangements = (arrangements of teachers) * (ways to select gaps) * (ways to permute students).Step 9: Thus, total arrangements = 6! * 7C6 * 6! = 6! * 7 * 6!.Step 10: Recognise that 7! = 7 * 6!, so 6! * 7 * 6! = 7! * 6!.
Verification / Alternative check:
We can sanity-check the formula by observing that there are enough teacher “separators” for the students. There are 7 gaps and only 6 students, so it is indeed possible to seat all students without adjacency. If we had 7 students instead, it would be impossible, as we would need to fill every gap, forcing two students to become adjacent when the circle closes. The final expression 7! * 6! exactly matches option 7! x 6!, confirming the correct combination of gap selection and student permutation.
Why Other Options Are Wrong:
- 7! x 7! suggests overcounting, effectively permuting students into 7 positions instead of 6, which is not needed and violates the constraint of having only 6 students.
- 6! x 6! ignores the 7 choices of which gap remains empty and thus undercounts the possible configurations.
- 7! x 5! similarly fails to respect the number of student permutations and gap selection structure, providing an incorrect factor.
Common Pitfalls:
Common mistakes include treating the arrangement as linear rather than circular, leading to an extra factor of 13 from rotations; or placing students first and then trying to insert teachers, which complicates the condition and often causes overcounting or invalid configurations. Forgetting that each gap can hold at most one student is another error that leads to arrangements where students are adjacent. Using the teachers as fixed separators and then assigning students to gaps systematically avoids these issues.
Final Answer:
The number of circular arrangements in which no two students sit together is 7! × 6!.
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