# Problem Analysis: Seating Arrangements with Restrictions
**Question**: In how many ways can 7 people sit around a round table if no two of the three specific people (Pierre, Rosa, and Thomas) can sit next to each other? (Rotations are considered identical.)
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## Attempt 1: Incorrect Approach
**Method**:
1. Seat the three restricted people (Pierre, Rosa, Thomas) in 4 chairs (non-adjacent).
- Calculation: \( \binom{4}{3} = 4 \) ways.
2. Seat the remaining 4 people in the remaining 4 chairs.
- Calculation: \( 7 \times 6 \times 5 \times 4 = 840 \).
3. Total: \( 4 \times 840 = 3360 \).
**Error**:
- Fails to account for circular permutations (rotations are equivalent).
- Overcounts by treating linear arrangements as distinct.
- **Final Answer**: ❌ Incorrect (3360).
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## Attempt 2: Partial Correction
**Method**:
1. Seat the three restricted people first, ensuring they are not adjacent.
- Assumes 3 valid configurations (unexplained).
2. Seat the remaining 4 people in \( 4! = 24 \) ways.
3. Seat the restricted people in 3 gaps between the 4 seated people.
- Calculation: \( 3 \times 24 \times 3 = 216 \).
**Error**:
- Incorrectly assumes only 3 valid configurations for the restricted people.
- Does not properly calculate gaps in circular arrangements.
- **Final Answer**: ❌ Incorrect (216).
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## Attempt 3: Correct Approach
**Method**:
1. **Step 1**: Arrange the 4 unrestricted people around the table.
- Circular permutations: \( (4-1)! = 6 \) ways.
- Creates 4 gaps between them.
2. **Step 2**: Choose 3 gaps out of 4 to place the restricted people.
- Combinations: \( \binom{4}{3} = 4 \).
3. **Step 3**: Arrange the 3 restricted people in the chosen gaps.
- Permutations: \( 3! = 6 \).
4. **Total**: \( 6 \times 4 \times 6 = 144 \).
**Validation**:
- Uses gap method to ensure no adjacency.
- Correctly applies circular permutation principles.
- **Final Answer**: ✅ **144**.
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## Conclusion
The correct number of valid seating arrangements is **144**. This is derived by:
1. Fixing the 4 unrestricted people to account for circular symmetry.
2. Placing the 3 restricted people in non-adjacent gaps.
3. Multiplying permutations for both groups.
**Final Answer**:
