Find how many distinct ways n objects can be arranged around a circle. Free, instant, and step-by-step — perfect for seating plans, arrangements, and combinatorics homework.
People around a table, beads on a ring, keys on a keyring — any n distinct objects arranged in a circle.
Enter a value and click Calculate to see the result.
How many different ways can five friends sit around a campfire? The answer is not 120 — it is 24. The moment objects are arranged in a circle instead of a row, something fundamental changes: rotating the whole arrangement does not create a new one. If everyone at the table shifts one seat clockwise, nobody has new neighbours, so for counting purposes nothing has changed. This single insight is the entire foundation of circular permutations, and it leads to one of the most elegant formulas in combinatorics: the number of distinct circular arrangements of n different objects is (n − 1)!. This free Circular Permutations Calculator applies that formula instantly, shows every step of the working, and explains why the answer is what it is — so whether you are planning a wedding seating chart, solving a combinatorics assignment, or designing a beaded bracelet, you get a correct, trustworthy result in seconds.
Using the calculator takes less than ten seconds, and each result comes with a full explanation so you can verify the mathematics yourself:
A note on the upper limit: factorials grow explosively — 18! is already 6,402,373,705,728,000 — and JavaScript numbers lose exact integer precision beyond 253. Capping n at 19 keeps every displayed digit exact, which matters when you are counting real arrangements rather than approximating them. If you need circular permutations for very large n, the formula (n − 1)! still applies; you would simply evaluate it with a big-integer library or a computer algebra system.
A permutation is an arrangement of objects in a definite order. When the objects stand in a row, counting is straightforward: the first position can be filled by any of the n objects, the second by any of the remaining n − 1, the third by any of the remaining n − 2, and so on, giving the famous product n × (n − 1) × (n − 2) × … × 1, written n! (n factorial). Five distinct books on a shelf can therefore be ordered in 5! = 120 ways.
A circular permutation asks the same question about a circle. At first glance you might expect the same n! answer, but the circle has a symmetry the row lacks: rotation. Imagine four people — A, B, C, D — seated clockwise around a table in that order. Now ask everyone to move one chair clockwise. The clockwise order is still A, B, C, D; every person still has exactly the same neighbours on the left and right. If what matters is who sits next to whom — which is the entire point of a seating arrangement — then these two seatings are the same arrangement, merely viewed from a different starting point. In a row, by contrast, shifting everyone one place changes who sits at each end, so the arrangements are genuinely different.
This is the key conceptual leap: in circular arrangements we consider two orderings identical when one can be rotated to match the other. Every genuinely distinct circular arrangement therefore corresponds to exactly n linear arrangements — the n rotations of it. Starting from the n! linear arrangements and grouping them into families of n rotations each, the number of distinct circular arrangements is n! ÷ n = (n − 1)!. Four people around a table give 4! ÷ 4 = 3! = 6 distinct seatings, not 24.
There is an equally intuitive way to see the same result, often called the “fix one” argument. Because rotations do not matter, you may arbitrarily declare that one particular object stays put — say, the guest of honour always sits in the “top” chair, or one specific bead is fixed at the clasp of the necklace. With one position pinned down, the circle is effectively “cut” at that point and straightened into a row of the remaining n − 1 objects, which can be ordered in (n − 1)! ways. Both arguments converge on the same formula, which is a good sign that the mathematics is sound.
One important refinement concerns flippable arrangements. A seating around a table is usually considered “directed”: clockwise order A-B-C-D is different from counter-clockwise order A-D-C-B, because you cannot flip the table over. But a necklace or bracelet can be turned over, so each arrangement and its mirror image are the same physical object. For such flippable circles the count is halved: (n − 1)! ÷ 2. This calculator reports the standard directed count (n − 1)!; if your objects can be flipped — beads, keys on a ring, charms on a bracelet — simply halve the result, provided n is greater than 2.
Pcircular(n) = (n − 1)!
where n is the number of distinct objects and ! denotes the factorial.
For flippable arrangements (necklaces, bracelets): P = (n − 1)! ÷ 2.
The formula rests on three conditions that are worth stating explicitly, because every common error in circular permutation problems comes from violating one of them. First, the objects must be distinct — five different people, not five identical chairs. If some objects are identical, you must divide further by the factorials of the repeat counts (see the FAQ below). Second, only relative order matters: arrangements that differ by rotation are counted once. If your problem treats rotations as different — for instance, numbered seats in a stadium — then you are counting linear permutations and should use n! instead. Third, the standard formula assumes the circle cannot be flipped; halve the result for necklaces and similar objects that can be turned over.
The derivation is worth memorising because it appears, in disguised form, across combinatorics. Begin with all n! linear orderings. Define an equivalence relation: two orderings are equivalent if one is a rotation of the other. Each equivalence class contains exactly n orderings (the n possible rotations — and they are all different as linear orderings because the objects are distinct). The number of classes is therefore n! ÷ n = (n − 1)!. Equivalently, pin one object to break the rotational symmetry and permute the rest: (n − 1)! ways. Both roads reach the same destination.
Example 1 — Five friends around a campfire. Five distinct people, rotations considered the same. Apply the formula: (5 − 1)! = 4! = 4 × 3 × 2 × 1 = 24 distinct seatings. Intuitively: fix Ana at the “north” spot, then arrange the other four friends in the remaining seats in 4! = 24 ways. Compare with the linear answer of 120 — the circle eliminates a factor of 5, exactly the five rotations of each seating.
Example 2 — Eight executives around a boardroom table, with the CEO and CFO required to sit together. Treat the CEO–CFO pair as a single fused block. We now have 7 objects (the block plus 6 others) to arrange circularly: (7 − 1)! = 6! = 720. Within the block, the two executives can sit in 2 orders (CEO-left-of-CFO or CFO-left-of-CEO), so multiply by 2! = 2. Total: 720 × 2 = 1,440 seatings. The “fuse into a block” trick handles most adjacency conditions in circular problems.
Example 3 — Seven distinct beads on a necklace. A necklace can be flipped over, so clockwise and counter-clockwise orders describe the same physical necklace. Start with the directed count (7 − 1)! = 6! = 720, then halve for the mirror symmetry: 720 ÷ 2 = 360 distinct necklaces. Forgetting this halving is the single most common error in necklace-type problems, so always ask yourself: can the arrangement be turned over?
Example 4 — Ten labelled keys on a keyring. A keyring, like a necklace, is flippable: (10 − 1)! ÷ 2 = 9! ÷ 2 = 362,880 ÷ 2 = 181,440 distinct orders. Notice how quickly the numbers explode — this is why the calculator caps n at 19, where (n − 1)! = 18! = 6,402,373,705,728,000, still exactly representable.
Circular permutations are far more than a textbook curiosity; they model any situation where only cyclic order matters. Event planners use them for wedding and banquet seating charts, where the goal is counting — or deliberately avoiding — particular neighbour pairings around round tables. Tournament organisers use circular arrangements to generate round-robin fixtures: seating n teams around a circle and rotating them produces each round’s pairings, a classical construction in sports scheduling. Jewellery designers count distinct bracelets and necklaces with the flippable variant (n − 1)! ÷ 2 before committing to production runs. Operations managers designing rotating shift rosters around a weekly cycle face the same mathematics: only the cyclic order of shifts matters, not where the “week” arbitrarily begins. In molecular biology, circular DNA plasmids are analysed up to rotation, so counting distinct arrangements of markers around a plasmid uses circular permutations. Even cryptography meets the idea when counting distinct rotations of a key schedule arranged on a ring. Whenever rotating the whole configuration changes nothing, (n − 1)! is the count you need.
Circular permutation entity summary: the number of distinct arrangements of n distinct objects around a circle, counting rotations as identical, is (n − 1)!. Derivation: n! linear permutations grouped into equivalence classes of n rotations each, giving n!/n = (n − 1)!; equivalently, fix one object and arrange the remaining n − 1 linearly. For arrangements that can also be flipped (necklaces, bracelets), the count is (n − 1)!/2. Requires distinct objects and unnumbered positions; with repeated objects divide by the factorials of repetition counts.
A circular permutation is an arrangement of objects around a circle in which rotations of the whole arrangement are considered identical. If everyone at a round table shifts one seat clockwise, each person keeps the same neighbours, so it counts as the same arrangement. For n distinct objects, the number of circular permutations is (n − 1)!.
In a row, all n! orderings are distinct. Around a circle, each distinct arrangement corresponds to n linear orderings — its n rotations — so we divide: n! ÷ n = (n − 1)!. Equivalently, you can “fix” one object in place to break the rotational symmetry and then arrange the remaining n − 1 objects in a line, giving (n − 1)! ways.
Using (n − 1)! with n = 6: (6 − 1)! = 5! = 5 × 4 × 3 × 2 × 1 = 120 distinct seatings. This assumes the seats are unnumbered and only who-sits-next-to-whom matters. If the chairs were numbered or assigned, rotations would count as different and the answer would be 6! = 720.
For n = 1, the formula gives (1 − 1)! = 0! = 1 — a single object has exactly one circular arrangement, which is correct. For n = 2, (2 − 1)! = 1! = 1: two people around a table always sit opposite each other, so there is only one distinct arrangement (each person has the other on both sides).
Linear permutations count arrangements in a row, where every position is distinguishable: n! arrangements. Circular permutations count arrangements around a circle, where rotating the entire arrangement changes nothing: (n − 1)! arrangements. The circle has n-fold rotational symmetry that the row lacks, so the count is smaller by exactly a factor of n.
Yes — when the arrangement can be flipped over. A necklace turned over shows the mirror image of its bead order, which is the same physical necklace, so each directed circular arrangement pairs with its mirror image. The count becomes (n − 1)! ÷ 2 for n > 2. Seating arrangements are not halved, because diners cannot be flipped through the table.
Start with the distinct-object count (n − 1)! and divide by the factorial of each repetition count. For example, arranging 7 beads around a necklace where 3 are identical red beads: (7 − 1)! ÷ 3! = 720 ÷ 6 = 120 directed arrangements (then halve to 60 if the necklace is flippable). Each set of identical objects overcounts by its internal permutations, hence the division.
With n = 10 distinct people around an unmarked round table: (10 − 1)! = 9! = 362,880 distinct seatings. If two particular people must sit together, fuse them into a block: arrange 9 objects circularly (8! = 40,320) and multiply by 2 for their internal order, giving 80,640 seatings.
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Last updated: October 9, 2026