In a circular arrangement there is no first or last position — only who sits next to whom. Turning everyone one seat to the left leaves every neighbor unchanged, so it is not a new arrangement. That single fact cuts the count from n! to (n − 1)!. This calculator handles full circles, circles that seat only some of the items, and necklaces or bracelets where flipping the circle over also changes nothing.
How to use the circular permutations calculator
- Enter Items available (n) — the people, beads or objects you can use.
- Leave Places in the circle (r) blank to place all n items, or enter a smaller number to seat only r of them.
- Under Which arrangements count as the same?, choose Rotations (round table) for seatings, or Rotations and flips (necklace) for objects that can be turned over.
- Read the count on the tape. The items list the line count and both circular counts side by side, and for up to 16 places a diagram shows one arrangement around the ring.
Circular permutation formulas
All n items around a table, rotations ignored:
Only r of the n items, seated in a circle of r places:
Necklaces and bracelets, where a mirror image is the same object (for r ≥ 3):
Worked examples
Eight guests at a round table. In a row: 8! = 40,320 orders. Each seating has 8 rotations, so 40,320 ÷ 8 = 5,040 = 7! distinct seatings.
Eight different beads on a bracelet. Turning the bracelet over pairs each arrangement with its mirror image: 5,040 ÷ 2 = 2,520.
Four of ten people at a small table. Choose and order 4 of 10: P(10, 4) = 10 × 9 × 8 × 7 = 5,040. Divide by the 4 rotations: 1,260.
Why fixing one item gives the same answer
Another way to see (n − 1)! is to seat one person first. Because the table has no head, it does not matter where they sit — every chair is equivalent. Once that person is seated, the remaining n − 1 chairs are now distinguishable (first to the left, second to the left, and so on), and the other guests can fill them in (n − 1)! ways. Both arguments agree, which is a good check.
Common variations
Someone must sit at the head. If one seat is special — a head chair, a window seat, a numbered place — rotations are no longer equivalent, and the count goes back to the line formula n!.
Couples sit together. Treat each couple as a single block, arrange the blocks around the table, then order each couple inside its block. Four couples give (4 − 1)! × 2⁴ = 6 × 16 = 96 seatings.
Two people must not sit together. Count all seatings and subtract the ones where they are adjacent. For 8 guests: 7! − 2 × 6! = 5,040 − 1,440 = 3,600.
Identical beads. If some beads share a color, rotations can map an arrangement onto itself, and simple division overcounts or undercounts. Problems like “how many necklaces with 3 red and 3 blue beads” need Burnside’s lemma rather than this calculator.
Line versus circle at a glance
| Items | In a row (n!) | Round table ((n − 1)!) | Necklace ((n − 1)! ÷ 2) |
|---|---|---|---|
| 3 | 6 | 2 | 1 |
| 4 | 24 | 6 | 3 |
| 5 | 120 | 24 | 12 |
| 6 | 720 | 120 | 60 |
| 8 | 40,320 | 5,040 | 2,520 |
| 10 | 3,628,800 | 362,880 | 181,440 |
For straight-line arrangements, use the permutations calculator; for the factorials themselves, the factorial calculator prints any n! up to 10,000! exactly.
Frequently asked questions
Why is the number of circular permutations (n − 1)!?
A row of n items can be arranged n! ways. Bending the row into a circle makes the n rotations of each arrangement look identical, so every circular arrangement was counted n times: n! ÷ n = (n − 1)!.
When should I divide by 2 as well?
Divide by 2 when an arrangement and its mirror image count as the same, as with a necklace or bracelet that can be turned over. Do not divide by 2 for people at a table, where clockwise and counterclockwise orders are different seatings.
What if the seats are numbered?
Then rotating everyone moves them to different seats, so rotations are new arrangements. Numbered or otherwise distinguishable seats make it an ordinary line problem: n! ways for n people.
Does the formula work for beads of the same color?
No. This calculator assumes every item is different. Counting necklaces with repeated colors needs Burnside's lemma or Pólya counting, because some colorings look the same after only part of a turn.
What happens with only one or two items?
One item has a single arrangement. Two items around a circle also have just one arrangement, and flipping it changes nothing, so the necklace rule does not divide by 2 when r is 1 or 2.