Circular Permutation Calculator
Count circular arrangements: (n − 1)! seatings around a round table, P(n, r)/r for part of a group, and ÷2 for necklaces and bracelets.
Circular permutations count the ways to arrange distinct objects around a circle when turning the whole circle does not create a new arrangement. Seating guests at a round table, threading beads on a bracelet or ordering players in a ring all use this count. The calculator handles arranging everyone or only some of the objects, and can treat mirror images as identical for necklaces and key rings.
How to use this tool
- Enter how many distinct objects (people, beads, keys) you have.
- Optionally enter how many of them go around the circle; leave it blank to use all of them.
- Tick the mirror-image box if the circle can be flipped over, like a bracelet.
- Read the number of circular arrangements and compare it with the count for a straight line.
Formula
All n objects: (n − 1)! r of n objects: P(n, r) / r = n! / ((n − r)! × r) Flips allowed: divide by 2 (when r ≥ 3)
In a line there are n! orders. Around a circle each order has n rotations that look the same, so the count is n! / n = (n − 1)!. If the circle can be turned over, each clockwise order matches its anticlockwise mirror image, which halves the count.
Example
8 guests at a round table: (8 − 1)! = 7! = 5,040 seatings, compared with 8! = 40,320 for a straight row. If those were 8 different beads on a bracelet that can be flipped, the count is 5,040 / 2 = 2,520. Seating 3 of 6 people at a round table for three gives 6 × 5 × 4 / 3 = 40.
Frequently asked questions
Why is it (n − 1)! and not n!?
Fix one person in place to remove the rotations. The remaining n − 1 people can then sit in (n − 1)! orders, and every one of those is genuinely different.
When should I divide by 2?
Only when clockwise and anticlockwise count as the same, which is true for objects you can pick up and turn over: necklaces, bracelets, key rings. People at a table are not flipped, so you do not halve.
Does this work with repeated objects?
No. The formula assumes every object is different. Circles with repeated colours, such as 3 red and 2 blue beads, need necklace counting with Burnside’s lemma instead.
What if the seats are numbered?
Then each rotation is a different seating, and the answer is the ordinary permutation count n!, or P(n, r) when only r seats are filled.
Last updated: 2026-10-30