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Module 4 · L5 of 15 ~35 min ⚡ +95 XP available

Circular Permutations

Eight friends sit down to dinner — how many different seating arrangements are there? If you answered $8!$, you'd be counting the same arrangement multiple times just because everyone shifted one seat clockwise. Circular permutations fix this by locking one person in place and counting only the relative arrangements of everyone else. It's one of those ideas that feels strange at first and obvious afterwards.

Today's hook — Six people sit around a circular table. You already know $6! = 720$ arrangements exist for a row. So why is the answer only $120$ for a circle? By the end of this lesson you'll understand exactly why we fix one person and divide — and you'll handle restriction problems without flinching.

0/5QUESTS

Recall — your gut answer first

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Six people are going to sit around a circular dinner table. Without using any formula — do you think the number of arrangements is the same as, less than, or more than the number of arrangements in a straight line? Write your prediction and reason before reading on.

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The core idea

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There is one key principle for circular permutations: fix one object to remove rotational equivalence, then arrange the rest. This gives $(n-1)!$ for $n$ distinct objects around a circle.

In a line, every position is distinct — shifting everyone one step gives a genuinely different arrangement. In a circle, rotating everyone one seat gives the same relative arrangement. Fixing one person's seat eliminates all $n$ rotational copies of each arrangement, leaving exactly $\dfrac{n!}{n} = (n-1)!$.

$(n-1)!$

Fix one object

Always fix one person or object in position before counting — this removes all $n$ rotational duplicates at once.

Numbered seats

If seats are labelled or otherwise distinguishable, use $n!$ — rotations create different arrangements when positions are fixed.

Restrictions still apply

Treat restricted units (e.g., two people who must sit together) as a single block, then multiply by the internal arrangements of that block.

What you'll master

Know

Key facts

Circular arrangements of $n$ distinct objects $= (n-1)!$
$n!$ applies when positions are fixed or labelled
Two objects sitting together: treat as one block, multiply by $2!$

Understand

Concepts

Why rotations of the same arrangement must not be counted separately
The relationship $\dfrac{n!}{n} = (n-1)!$
How fixing one object eliminates rotational duplicates

Can do

Skills

Count circular arrangements for $n$ objects
Apply restrictions (must sit together / must not sit together)
Distinguish between circular and linear problems under exam conditions

Key terms

Circular permutationAn arrangement of objects around a circle where only relative positions matter; rotations of the same arrangement are identical.

Linear permutationAn arrangement in a row where every position is distinct and $n!$ arrangements exist for $n$ objects.

Rotational equivalenceTwo circular arrangements are equivalent if one is a rotation of the other; only one is counted.

Fix one objectHolding one object in a designated position to provide a reference frame, eliminating all $n$ rotational duplicates.

Restriction (together)Two objects that must be adjacent are grouped as one block, reducing $n$ objects to $n-1$ units for the circular count.

Reflection symmetryClockwise and anticlockwise arrangements are treated as different unless the problem states the circle is flipped (e.g., a necklace).

Circular arrangements — the formula and why it works

core concept

When $n$ distinct objects are arranged in a straight line there are $n!$ arrangements because every position is uniquely defined. In a circle, no position is inherently "first" — rotating everyone one seat produces the same relative arrangement.

To count distinct circular arrangements, fix one object in an arbitrary position. This removes the rotational freedom. The remaining $n-1$ objects can be arranged in the remaining $n-1$ seats in $(n-1)!$ ways.

$$\text{Circular arrangements of } n \text{ distinct objects} = (n-1)!$$

Alternatively: there are $n!$ linear arrangements. Each circular arrangement is produced by $n$ rotations of itself, so the number of distinct circular arrangements is $\dfrac{n!}{n} = (n-1)!$.

Real-world connection. Round-table diplomacy: the number of ways 5 world leaders can be seated around a circular table is $(5-1)! = 24$. If they sat in a row it would be $5! = 120$ — five times more, because each of the 5 rotations of a circular arrangement would count as a distinct row arrangement.

Circular permutations of n distinct objects = (n-1)!; Reason: fix one object to remove rotational equivalence; arrange remaining n-1 objects in (n-1)! ways

Pause — copy the circular permutation formula into your book: $n$ distinct objects around an unlabelled circle $= (n-1)!$; reason — fix one object to remove rotational equivalence, arrange the remaining $n-1$ linearly.

Quick check: In how many ways can 6 people be seated around a circular table?

Circular vs linear — when to use which

core concept

We just saw that circular permutations of $n$ distinct objects $= (n-1)!$ because fixing one object removes the rotational equivalences. That raises a question: when is a table arrangement treated as circular $(n-1)!$ rather than linear $n!$ — and what if the seats themselves are distinguishable? This card answers it → use $(n-1)!$ for an unlabelled circle; use $n!$ if seats are labelled or fixed (e.g. chairs with nameplates).

The choice between $(n-1)!$ and $n!$ depends entirely on whether positions are distinguishable:

Relative positions only (unlabelled round table, circle of friends): use $(n-1)!$. Two arrangements that are rotations of each other are the same.
Fixed positions (numbered seats, chairs bolted down, seats on a Ferris wheel with a sign): use $n!$. Rotating everyone gives a genuinely different assignment of people to seats.

A useful test: if I rotate everyone one position clockwise, does this produce a new arrangement or the same one? If it's the same, you have a circular problem.

Clockwise vs anticlockwise. Unless the problem says otherwise (e.g., "a necklace can be flipped"), treat clockwise and anticlockwise as distinct arrangements. For objects on a necklace (where flipping is allowed), divide by 2: $\dfrac{(n-1)!}{2}$.

Unlabelled circle (n-1)!; labelled/fixed positions n!; Test: does rotating everyone produce a new arrangement? No circular formula

Pause — copy the circular vs linear decision rule into your book: unlabelled circle → $(n-1)!$; labelled/fixed positions → $n!$; test: does rotating everyone produce a new arrangement? If yes, $(n-1)!$.

Did you get this? True or false: if 8 people sit on 8 numbered chairs arranged in a circle, the number of arrangements is $8!$.

Worked examples · 3 in a row, reveal as you go

PROBLEM 1 · BASIC CIRCULAR

In how many ways can 6 people be seated around a circular table?

Positions are not labelled — rotations are equivalent. Use the circular formula.

The table has no head; only relative positions matter.

PROBLEM 2 · RESTRICTION: MUST SIT TOGETHER

In how many ways can 5 people be seated around a circular table if two particular people, Alice and Bob, must sit together?

Treat Alice and Bob as a single unit. There are now $5 - 2 + 1 = 4$ units around the circle.

Grouping adjacent objects into a block is the standard restriction strategy.

PROBLEM 3 · RESTRICTION: MUST NOT SIT TOGETHER

In how many ways can 6 people be seated around a circular table if two particular people, Alice and Bob, must not sit together?

Total unrestricted circular arrangements of 6 people $= (6-1)! = 5! = 120$

Find the total first, then subtract the restricted cases.

Fill the gap: 4 boys and 4 girls alternate around a circle. First fix one girl; arrange the remaining 3 girls in $3!$ ways; arrange the 4 boys in the 4 gaps in $4!$ ways. Total $= 3! \times 4! = $ .

Misconceptions to fix · the 3 traps that cost marks

Trap 01

Using $n!$ for a circle

Writing $n!$ when seats are unlabelled counts each arrangement $n$ times — once for every rotation. Always ask: are positions distinguishable? If not, use $(n-1)!$.

Trap 02

Forgetting internal block arrangements

When two people are grouped into a block, their internal order matters. If Alice–Bob and Bob–Alice are different seatings, multiply by $2!$. Don't forget this factor.

Trap 03

Treating clockwise = anticlockwise

For standard table problems, clockwise and anticlockwise are different. Only divide by 2 (necklace formula) when the problem explicitly states the circle can be flipped or reversed.

Did you get this? True or false: the number of ways to arrange a necklace with 6 distinct beads (where flipping is allowed) is $\dfrac{5!}{2} = 60$.

Key formulas summary

$$\text{Circular arrangements of } n \text{ objects} = (n-1)!$$

$$\text{Necklace (flippable): } \frac{(n-1)!}{2}$$

$$\text{Together: } (n-1-1)! \times (\text{internal block arrangements})$$

$$\text{Not together: } (n-1)! - \text{(together count)}$$

Work mode · how are you completing this lesson?

Activities · practice with the ideas

Odd one out: Which of the following does not use a circular permutation formula?

In how many ways can 8 people be seated around a circular table?

In how many ways can 4 boys and 4 girls be seated around a circular table if they must alternate?

In how many ways can 6 people be seated around a circular table if two particular people must not sit together?

Seven people sit at a round table. In how many ways can they sit if three particular people must all sit together?

Explain in your own words why arranging $n$ objects around a circle gives $(n-1)!$ arrangements, not $n!$.

Revisit your thinking

Earlier you were asked: are circular arrangements fewer than, equal to, or more than linear arrangements?

The answer: always fewer. For $n$ objects, linear arrangements $= n!$ and circular arrangements $= (n-1)! = \dfrac{n!}{n}$. So circular arrangements are exactly $\dfrac{1}{n}$ of linear arrangements. For $n=6$: linear $= 720$, circular $= 120$ — exactly $\dfrac{1}{6}$ of 720.

If circular permutations are still unclear, draw diagrams of $n = 3$ (3 people A, B, C) and physically rotate the arrangements to see that there are only $(3-1)! = 2$ distinct seatings: A–B–C and A–C–B (going clockwise).

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Multiple choice

+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 41 mark

Q1. In how many ways can 8 people be seated around a circular table? (1 mark)

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ApplyBand 42 marks

Q2. In how many ways can 4 boys and 4 girls be seated around a circular table if they must alternate? (2 marks)

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AnalyseBand 52 marks

Q3. In how many ways can 6 people be seated around a circular table if two particular people must not sit together? (2 marks)

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Comprehensive answers (click to reveal)

Activity 1: 1. $(8-1)! = 7! = 5040$ · 2. Fix one girl: $3!$ ways for remaining girls; $4!$ ways for boys in 4 gaps. Total $= 3! \times 4! = 6 \times 24 = 144$ · 3. Total $= 5! = 120$; together $= (4-1)! \times 2! = 6 \times 2 = 12$; not together $= 108$ · 4. Block of 3 gives 5 units: $(5-1)! \times 3! = 24 \times 6 = 144$ · 5. In a circle, $n$ rotations of each linear arrangement all look the same, so we divide by $n$: $n!/n = (n-1)!$.

Q1 (1 mark): $(8-1)! = 7! = 5040$ [1].

Q2 (2 marks): Fix one girl in position [0.5]. Remaining 3 girls can be seated in $3! = 6$ ways [0.5]. The 4 boys fill the 4 gaps between girls in $4! = 24$ ways [0.5]. Total $= 6 \times 24 = 144$ [0.5].

Q3 (2 marks): Total circular arrangements $= (6-1)! = 120$ [0.5]. Treat the two together as a block: $(4-1)! \times 2! = 3! \times 2 = 12$ arrangements where they sit together [1]. Not together $= 120 - 12 = 108$ [0.5].

Boss battle · The Circle Master

earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena

Science Jump · platform challenge

Climb platforms by answering combinatorics questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

← Lesson 4 · Permutations with Restrictions Lesson 6 · Combinations →

Module overview · Extension 1 · Checkpoint 1