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.
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.
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 $\dfrac{n!}{n} = (n-1)!$.
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.
Alternatively: there are $n!$ linear arrangements. Each circular arrangement is produced by $n$ rotations of itself, so the number of distinct circular arrangements is $\dfrac{n!}{n} = (n-1)!$.
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?
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:
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.
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
In how many ways can 6 people be seated around a circular table?
In how many ways can 5 people be seated around a circular table if two particular people, Alice and Bob, must 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?
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
Did you get this? True or false: the number of ways to arrange a necklace with 6 distinct beads (where flipping is allowed) is $\dfrac{5!}{2} = 60$.
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 $n!$.
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).
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. In how many ways can 8 people be seated around a circular table? (1 mark)
Q2. In how many ways can 4 boys and 4 girls be seated around a circular table if they must alternate? (2 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)
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].
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