Permutations and circular arrangements
Use factorials for permutations and distinguish circular orders up to rotation from arrangements in labeled seats.
Content updated: October 22, 2026
Permutations count arrangements of all the distinct elements. For arrangements around a circle, the count depends on whether positions are distinguished or arrangements differing only by rotation are identified.
Learning goals: use factorials and distinguish rows, circular order, and numbered seats.
Prerequisites: the multiplication principle (English lesson); the difference between an order and a set.
On this page: The idea · Method · Examples · Practice · Further learning
1. Permutations of distinct elements
Let $A$ be a set of $n$ distinct elements, with $n\ge1$. A permutation is an ordered list containing each element of $A$ exactly once. Write $P_n$ for the number of permutations.
There are $n$ choices for the first element, $n-1$ for the second, and so on until the last position has $1$ choice. By the multiplication principle,
Each list uniquely determines the choices by position, so no arrangement is repeated or omitted. For the empty set, there is exactly one empty list; the convention $0!=1$ corresponds to this case. If some objects are regarded as identical, the distinctness hypothesis fails and $n!$ cannot be applied directly.
2. When do circular permutations apply?
For circular permutations, arrangements differing only by a rotation of the entire circle are identified. Fix a direction of reading around the circle; reflection, which reverses the order, is not identified. No numbered seat or external position distinguishes any seat.
Fix a particular person $A$ as a reference. Each circular order has exactly one representation beginning with $A$ in the chosen direction; the remaining $n-1$ people are then arranged in an order. Thus for $n\ge1$, the number of circular permutations is
This argument also holds for $n=1$, yielding $0!=1$. If seats are labeled or positions distinguished, moving $A$ to two different seats produces different outcomes; there are then $n!$ arrangements even if the seats form a circle. “Identical seats” leads to the circular formula only if rotations of the entire arrangement are also identified.
3. Worked examples
Example 1. Three people in a row and around a circle
Three friends $A,B,C$ can occupy three positions in a row in $3!=6$ ways. Read from left to right, these are
ABC, ACB, BAC, BCA, CAB, CBA.
If three seats are equally spaced around a circle and whole-circle rotations are identified, a clockwise rotation through $120^\circ$ does not create a new circular order. The six lists form two rotation classes: $ABC,BCA,CAB$ and $ACB,CBA,BAC$.
Fix $A$ as the starting point and read clockwise. The two representatives are $ABC$ and $ACB$, so there are $2!=(3-1)!=2$ orders. Reverse reading orders are not automatically identified by reflection.
Example 2. Four people with only circular order distinguished
Arrange $A,B,C,D$ around four unnumbered seats, identifying rotations. Fix $A$ and arrange $B,C,D$ clockwise. There are $3!=6$ ways. Fixing $A$ merely selects one representative of each circular order; it does not add a labeled seat to the problem.
Example 3. Four numbered seats
Use the same four people, but number the seats from $1$ to $4$. Placing $A$ in seat $1$ differs from placing $A$ in seat $2$. Filling seats $1,2,3,4$ successively allows $4,3,2,1$ choices, respectively, so the total is
Alternatively, take the $6$ circular orders in Example 2 and choose one of $4$ seats for $A$: $6\cdot4=24$. This determines the other three seats uniquely from the chosen order.
4. Practice exercises
- In how many ways can $5$ distinct students occupy $5$ seats in a row?
- Arrange $5$ people around a round table, identifying rotations but distinguishing the two reading directions. How many arrangements are there?
- Arrange $5$ distinct people in a row with $A,B$ adjacent. How many arrangements are there?
Show answers and hints
- $5!=120$.
- Fix one person as a reference; there are $4!=24$ arrangements.
- Treat $A,B$ as one block: there are $4!$ orders of the four objects and $2$ internal block orders. The total is $2\cdot4!=48$. Each valid row has this unique block, so no arrangement is counted twice.
5. Common mistakes
- Using $(n-1)!$ merely because the table is round, without checking whether seats are labeled.
- Dividing again by $2$ even though the problem distinguishes reverse orders.
- Using $n!$ when selecting only some of the $n$ elements.
- Forgetting that elements must be distinct or allowing a person to occur more than once.
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