🔄 Circular Permutations

Arranging objects in a circle with considerations for rotation and reflection symmetry.

🎯 Recognition Cues

  • Keywords: “around a table”, “in a circle”, “on a ring”, “circular”, “round”
  • Setup: Objects arranged in a circular pattern
  • Constraints: Rotation symmetry, reflection symmetry, fixed positions

📋 Solution Template

  1. Identify the symmetry type:

    • Rotation only: Use $(n-1)!$ formula
    • Rotation + reflection: Use $\frac{(n-1)!}{2}$ formula
    • Fixed positions: Fix some elements, arrange the rest

  2. Apply the appropriate method:

    • Basic circular: $(n-1)!$ (fix one element, arrange the rest)
    • With reflection: $\frac{(n-1)!}{2}$ (divide by 2 for reflection symmetry)
    • Mixed constraints: Combine with other counting techniques

  3. Check for AMC conventions:

    • People around table: Usually rotation only
    • Beads on bracelet: Usually rotation + reflection
    • Fixed positions: May reduce the problem

💡 Micro-Examples

Basic Circular

  • Problem: How many ways can you seat 4 people around a round table?
  • Solution: $(4-1)! = 3! = 6$ ways

With Reflection

  • Problem: How many ways can you arrange 4 beads on a bracelet?
  • Solution: $\frac{(4-1)!}{2} = \frac{3!}{2} = 3$ ways

Fixed Positions

  • Problem: How many ways can you seat 5 people around a round table with A in a fixed position?
  • Solution: Fix A, arrange the rest: $(5-1)! = 4! = 24$ ways

⚠️ Common Pitfalls & Variants

Pitfall: Using $n!$ instead of $(n-1)!$

  • Wrong: “Seat 4 people around a table” = $4! = 24$ ways
  • Right: $(4-1)! = 6$ ways (fix one person, arrange the rest)

Pitfall: Forgetting reflection symmetry

  • Wrong: “Arrange 4 beads on a bracelet” = $(4-1)! = 6$ ways
  • Right: $\frac{(4-1)!}{2} = 3$ ways (divide by 2 for reflection)

Pitfall: Confusing circular with linear arrangements

  • Wrong: “Arrange 4 people in a line” = $(4-1)! = 6$ ways
  • Right: $4! = 24$ ways (linear arrangement)

🏆 AMC-Style Worked Example

Problem: How many ways can you seat 6 people around a round table so that no two men sit together, given that there are 3 men and 3 women?

Solution:

  • Step 1: Arrange the 3 women around the table: $(3-1)! = 2! = 2$ ways
  • Step 2: This creates 3 gaps between women for the men
  • Step 3: Place the 3 men in these 3 gaps: $3! = 6$ ways
  • Step 4: Total: $2 \cdot 6 = 12$ ways

Key insight: Use the gaps method for spacing constraints in circular arrangements!

🔗 Related Topics

Next: Symmetry Probability → Hypergeo vs Binomial