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Learn circular permutation formulas for seating arrangements and circular problems. Learn why rotations are identical with worked examples and practice.

---

## 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 $$n$$ different objects in circular formation, we use the formula:

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

```math
P_n = (n-1)!
```

Where:

- $$P_n$$ = circular permutation of $$n$$ objects
- $$n$$ = number of objects to be arranged
- $$(n-1)!$$ = factorial of $$(n-1)$$

Visible text: - = circular permutation of objects
- = number of objects to be arranged
- = factorial of

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

Visible text: **Why is the formula and not ?**

The key concept is that rotation does not change a circular arrangement. Let's see it with a round-table example.

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.

Visible text: Imagine children (, , ) sitting around a round table. The arrangements , , and 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: $$n-1$$
4. Number of ways: $$(n-1)!$$

Visible text: 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: 
4. Number of ways:

For $$3$$ children: $$P_3 = (3-1)! = 2! = 2 \times 1 = 2 \text{ ways}$$.

Visible text: For children: .

## 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:

```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:

```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.

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

**Solution strategy:**

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

Visible text: 1. **Group** each couple as one unit → 
2. **Arrange these units** in a circle: 
3. **Arrange positions** within each couple: per couple
4. **Total calculation:**

## 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. $$5$$ 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. $$7$$ 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?

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

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

3. 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. 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: $$120 \text{ ways}$$**

   Solution steps:
   - Given: $$n = 6 \text{ people}$$
   - Circular permutation formula: $$P_n = (n-1)!$$
   - $$P_6 = (6-1)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$

   Therefore, $$6 \text{ people}$$ can sit around a campfire in $$120$$ different ways.

2. **Answer: $$5{,}040 \text{ ways}$$**

   Solution steps:
   - Given: $$8$$ different beads will be arranged in a circle
   - Circular permutation formula: $$P_n = (n-1)!$$
   - Calculation: $$P_8 = (8-1)! = 7!$$
   - $$7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5.040$$

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

3. **Answer: $$768 \text{ ways}$$**

   Solution steps:
   - Given: $$5$$ married couples ($$10 \text{ people}$$), each couple must be adjacent
   - **Grouping technique:** Consider each couple as one unit → $$5 \text{ units}$$
   - Circular permutation of $$5 \text{ units}$$: $$(5-1)! = 4! = 4 \times 3 \times 2 \times 1 = 24$$
   - Each couple can exchange positions: $$2! = 2 \text{ ways}$$ per couple
   - Total: $$24 \times 2^5 = 24 \times 32 = 768$$

   There are $$768$$ seating arrangements that meet the conditions.

4. **Answer: $$480 \text{ ways}$$**

   Solution steps (**complement method**):
   - Total ways without restrictions: $$(7-1)! = 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720$$
   - To count the unwanted ways, make the $$2 \text{ students}$$ who sit adjacent into one unit.
   - Now there are $$6$$ units around the circle, so the number of circular arrangements is $$(6-1)! = 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$$.
   - The $$2 \text{ students}$$ inside that unit can exchange positions, so there are $$2! = 2$$ internal arrangements.
   - Total adjacent arrangements: $$120 \times 2 = 240$$
   - **Desired ways:** $$720 - 240 = 480$$

   Therefore, there are $$480 \text{ ways}$$ to form a circle where the two students are not adjacent.

Visible text: 1. **Answer: **

 Solution steps:
 - Given: 
 - Circular permutation formula: 
 - 

 Therefore, can sit around a campfire in different ways.

2. **Answer: **

 Solution steps:
 - Given: different beads will be arranged in a circle
 - Circular permutation formula: 
 - Calculation: 
 - 

 The bracelet can be made in different arrangements.

3. **Answer: **

 Solution steps:
 - Given: married couples (), each couple must be adjacent
 - **Grouping technique:** Consider each couple as one unit → 
 - Circular permutation of : 
 - Each couple can exchange positions: per couple
 - Total: 

 There are seating arrangements that meet the conditions.

4. **Answer: **

 Solution steps (**complement method**):
 - Total ways without restrictions: 
 - To count the unwanted ways, make the who sit adjacent into one unit.
 - Now there are units around the circle, so the number of circular arrangements is .
 - The inside that unit can exchange positions, so there are internal arrangements.
 - Total adjacent arrangements: 
 - **Desired ways:** 

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