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Combinatorics

Circular Permutation

Understanding Circular Permutation

Have you ever sat with friends around a round table? Or played traditional games that form a circle? Such situations involve the concept of circular permutation.

Circular permutation is the arrangement of objects arranged around a circle. Unlike regular permutation which is arranged in a straight line, circular permutation considers the relative positions between objects in circular formation.

Why is it called circular? Because in circular arrangements, there is no fixed starting or ending position. Each object can serve as a reference point, so several different arrangements in a straight line can be considered the same in circular arrangement.

Circular Permutation Formula

To determine the number of ways to arrange nn different objects in circular formation, we use the formula:

Pn=(n1)!P_n = (n-1)!

Where:

  • PnP_n = circular permutation of nn objects
  • nn = number of objects to be arranged
  • (n1)!(n-1)! = factorial of (n1)(n-1)

Why is the formula (n1)!(n-1)! and not n!n!?

Key concept: in circular arrangements, rotation does not change the arrangement. Let's understand this with an analogy:

Imagine 3 children (A, B, C) sitting around a round table. The arrangements ABC, BCA, and CAB are actually the same arrangement when viewed from a circular perspective, because their relative positions remain unchanged.

Calculation steps:

  1. Fix one object as a reference point (for example, child A)
  2. Arrange other objects relative to this reference point
  3. Remaining objects to be arranged: n1n-1
  4. Number of ways: (n1)!(n-1)!

For 3 children: P3=(31)!=2!=2×1=2P_3 = (3-1)! = 2! = 2 \times 1 = 2 ways.

Applications in Daily Life

Circular permutation is often encountered in various real situations:

Circular Seating:

Five students will sit around a round table for discussion. The number of ways they can sit is:

P5=(51)!=4!=4×3×2×1=24 waysP_5 = (5-1)! = 4! = 4 \times 3 \times 2 \times 1 = 24 \text{ ways}

Traditional Games:

Eight children play in a circle. The number of different formations they can form is:

P8=(81)!=7!=7×6×5×4×3×2×1=5.040 waysP_8 = (8-1)! = 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5.040 \text{ ways}

Situations with Special Conditions:

When there are additional conditions such as certain objects must be adjacent, we use grouping technique:

Example: 4 married couples sit in a circle, each couple must be adjacent.

Solution strategy:

  1. Group each couple as one unit → 4 units
  2. Arrange these units in a circle: (41)!=3!=6(4-1)! = 3! = 6 ways
  3. Arrange positions within each couple: 2!2! ways per couple
  4. Total calculation: 3!×(2!)4=6×24=6×16=963! \times (2!)^4 = 6 \times 2^4 = 6 \times 16 = 96 ways

Practice Problems

  1. There are 6 friends who will sit around a campfire. How many ways can they sit?

  2. A bracelet will be made from 8 different colored beads. How many ways can the beads be arranged on the bracelet?

  3. Five married couples will sit around a round table with the condition that each husband must sit next to his wife. How many possible seating arrangements are there?

  4. Seven students will play a circular game, but two specific students must not sit adjacent to each other. How many ways can they form a circle?

Answer Key

  1. Answer: 120 ways

    Solution steps:

    • Given: n=6n = 6 people
    • Circular permutation formula: Pn=(n1)!P_n = (n-1)!
    • P6=(61)!=5!=5×4×3×2×1=120P_6 = (6-1)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120

    Therefore, 6 people can sit around a campfire in 120 different ways.

  2. Answer: 5,040 ways

    Solution steps:

    • Given: 8 different beads will be arranged in a circle
    • Circular permutation formula: Pn=(n1)!P_n = (n-1)!
    • Calculation: P8=(81)!=7!P_8 = (8-1)! = 7!
    • 7!=7×6×5×4×3×2×1=5.0407! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5.040

    The bracelet can be made in 5,040 different arrangements.

  3. Answer: 768 ways

    Solution steps:

    • Given: 5 married couples (10 people), each couple must be adjacent
    • Grouping technique: Consider each couple as one unit → 5 units
    • Circular permutation of 5 units: (51)!=4!=4×3×2×1=24(5-1)! = 4! = 4 \times 3 \times 2 \times 1 = 24
    • Each couple can exchange positions: 2!=22! = 2 ways per couple
    • Total: 24×25=24×32=76824 \times 2^5 = 24 \times 32 = 768

    There are 768 seating arrangements that meet the conditions.

  4. Answer: 480 ways

    Solution steps (complement method):

    • Total ways without restrictions: (71)!=6!=6×5×4×3×2×1=720(7-1)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720
    • Count unwanted ways (2 students sitting adjacent):
      • Consider 2 students as one unit: (61)!=5!=5×4×3×2×1=120(6-1)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120
      • 2 students can exchange positions: 2!=22! = 2
      • Total adjacent arrangements: 120×2=240120 \times 2 = 240
    • Desired ways: 720240=480720 - 240 = 480

    Therefore, there are 480 ways to form a circle where the two students are not adjacent.