Circular Permutation

Rotations of the Same Circular Arrangement

At a round table, moving every person one seat clockwise does not change who sits next to whom. A circular permutation therefore counts each relative seating order once. Rotations of the same order belong to the same arrangement.

This differs from a linear permutation. In a row, the first and last positions are fixed. In a circle, there is no distinguished starting position.

Any object can therefore serve as the reference point. Linear orders that differ only by a rotation represent the same circular arrangement.

Circular Permutation Formula

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

Where:

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

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

Rotating everyone together does not create a new circular arrangement. The round-table example below shows why.

Suppose $3$ children, $A$, $B$, and $C$, sit around a round table. The orders $ABC$, $BCA$, and $CAB$ are the same arrangement because each is a rotation of the others and the relative positions do not change.

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: $(n-1)!$

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

The formula $(n-1)!$ identifies rotations but still treats mirror images as different. Treating mirror images as different is appropriate for people facing the center of a table because reversing the order swaps left and right neighbors. A bracelet can also be flipped. For $n\geq3$ distinct beads, every circular order pairs with a different mirror image, so the bracelet has $\frac{(n-1)!}{2}$ arrangements.

Counting Circular Arrangements

The same formula counts circular seating orders and other formations arranged around a center.

Circular Seating:

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

Traditional Games:

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

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 → $4 \text{ units}$
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: $3! \times (2!)^4 = 6 \times 2^4 = 6 \times 16 = 96 \text{ ways}$

Practice Problems

The problems differ in one condition each: a plain circle, a bracelet where mirror images match, and couples who must sit together. Pick the matching formula before calculating.

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

2. A bracelet will be made from $8$ beads of different colors. If rotations and flipped mirror images count as the same bracelet, how many arrangements are possible?

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?

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: $2\,520 \text{ ways}$

   Solution steps:

   • First count rotations as identical: $P_8 = (8-1)! = 7!$
   • $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5\,040$
   • Flipping the bracelet makes each clockwise order identical to its counterclockwise mirror image.
   • Because all beads are different, these arrangements form pairs, so divide by $2$.
   • $\frac{5\,040}{2} = 2\,520$

   The bracelet can be made in $2\,520$ 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.