# Nakafa Framework: LLM

URL: /en/subject/high-school/12/mathematics/combinatorics/circular-permutation
Source: https://raw.githubusercontent.com/nakafaai/nakafa.com/refs/heads/main/packages/contents/subject/high-school/12/mathematics/combinatorics/circular-permutation/en.mdx

Output docs content for large language models.

---

export const metadata = {
   title: "Circular Permutation",
   description: "Master circular permutation formula (n-1)! for seating arrangements & circular problems. Learn why rotations are identical with step-by-step examples & practice.",
   authors: [{ name: "Nabil Akbarazzima Fatih" }],
   date: "05/26/2025",
   subject: "Combinatorics",
};

## 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 <InlineMath math="n" /> different objects in circular formation, we use the formula:

<BlockMath math="P_n = (n-1)!" />

Where:
- <InlineMath math="P_n" /> = circular permutation of <InlineMath math="n" /> objects
- <InlineMath math="n" /> = number of objects to be arranged
- <InlineMath math="(n-1)!" /> = factorial of <InlineMath math="(n-1)" />

**Why is the formula <InlineMath math="(n-1)!" /> and not <InlineMath math="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: <InlineMath math="n-1" />
4. Number of ways: <InlineMath math="(n-1)!" />

For 3 children: <InlineMath math="P_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:

<BlockMath math="P_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:

<BlockMath math="P_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: <InlineMath math="(4-1)! = 3! = 6" /> ways
3. **Arrange positions** within each couple: <InlineMath math="2!" /> ways per couple
4. **Total calculation:** <InlineMath math="3! \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: <InlineMath math="n = 6" /> people
   - Circular permutation formula: <InlineMath math="P_n = (n-1)!" />
   - <InlineMath math="P_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: <InlineMath math="P_n = (n-1)!" />
   - Calculation: <InlineMath math="P_8 = (8-1)! = 7!" />
   - <InlineMath math="7! = 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: <InlineMath math="(5-1)! = 4! = 4 \times 3 \times 2 \times 1 = 24" />
   - Each couple can exchange positions: <InlineMath math="2! = 2" /> ways per couple
   - Total: <InlineMath math="24 \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: <InlineMath math="(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: <InlineMath math="(6-1)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120" />
     - 2 students can exchange positions: <InlineMath math="2! = 2" />
     - Total adjacent arrangements: <InlineMath math="120 \times 2 = 240" />
   - **Desired ways:** <InlineMath math="720 - 240 = 480" />
   
   Therefore, there are 480 ways to form a circle where the two students are not adjacent.