Quantitative Aptitude · Permutations and Combinations

Circular Permutations for CA Foundation

Updated 1 October 2026 · Fact-checked

A circular permutation counts ways to arrange n distinct objects around a circle, where only relative positions matter. Fix one object to remove rotations and arrange the rest: (n − 1)!. For necklaces or garlands, which can be flipped over, divide by 2: (n − 1)! ÷ 2.

Understand Circular Permutations

In a line, ABC and BCA are different arrangements. Around a circle, they are the same if everyone just moves one seat along. Rotating the whole circle does not create a new arrangement. That is the key idea.

Take n distinct objects. Placed in a line, they give n! arrangements. Around a circle, each circular arrangement shows up n times in that count, once for each rotation. So the number of circular arrangements is n! ÷ n = (n − 1)!.

An easy way to see it: fix one person at a seat as a reference point. Now the remaining n − 1 people can be placed in the other seats in (n − 1)! ways.

Some circles can be turned over. A necklace or a garland looks the same from the front and the back, so a clockwise order and its anticlockwise mirror are the same arrangement. Here you divide by 2, giving (n − 1)! ÷ 2. This holds when n ≥ 3 and all objects are distinct.

People seated at a round table are different. Clockwise and anticlockwise orders are different arrangements, since you cannot flip people over. So use (n − 1)!.

Key formulas to remember

Circular arrangement of n distinct objects (clockwise and anticlockwise different): (n − 1)!; Use for round tables, people in a ring, and similar cases.
Necklace or garland (clockwise and anticlockwise same): (n − 1)! ÷ 2; Use for n ≥ 3 distinct beads or flowers that can be flipped.
Selecting r from n and arranging in a circle: nCr × (r − 1)! = nPr ÷ r; Dividing nPr by r removes the r rotations.
Two particular persons always together: (n − 2)! × 2!; Treat the pair as one unit: n − 1 units give (n − 2)! circular ways, times 2! for the pair's order.
Two particular persons not together: (n − 1)! − (n − 2)! × 2!; Total minus together.
Women or men placed in gaps: (m − 1)! × mPk; Seat m people in a circle first in (m − 1)! ways. This creates m gaps. Placing k others in these gaps, one per gap, gives mPk ways. When k = m (equal groups, alternate seating), mPk becomes m!, so the total is (m − 1)! × m!.

How to solve Circular Permutations questions

Follow these steps for almost any circular arrangement question.

1Check whether the objects are distinct. If some are identical, adjust by dividing by the factorials of repeated items.
2Decide if the arrangement can be flipped. Round table or seating means clockwise and anticlockwise differ. Necklace, garland or bracelet means they are the same.
3If you must select first, choose with nCr, then arrange the chosen items in a circle.
4Handle restrictions by grouping or by gaps. For 'together', make a block. For 'alternate', seat one group first, then the other in the gaps.
5Fix one object as the reference point so rotations are removed. When there are restrictions, fix the object that is part of the restriction.
6Write the total as a product of factorials, such as (n − 1)!, and apply the ÷ 2 only for flippable arrangements.
7Evaluate the number and match it with the options.

Quickest way: Reduce to (n − 1)! and scan the options

When to use it: Use this in the MCQ paper when the question is a direct seating or necklace problem.

Ask one question: can it be flipped over? If yes, divide by 2.
Write the answer in factorial form first. Do not expand if the options are in factorial form.
For 'together', count the units: with the block, n − 1 units give (n − 2)! circular ways, then multiply by the block's internal order.
For men and women alternate with equal numbers (m of each), seat one gender in (m − 1)! ways, then arrange the other gender in the m gaps in m! ways. If the second group has k people and k is not m, use mPk instead.
When the question mentions a necklace, watch for options equal to n! or (n − 1)!. These are common distractors that ignore the flip.
If the arrangement has many restrictions and the numbers are large, skip and return later.

Common mistakes in Circular Permutations

Using n! for a round table.
Students carry over the linear rule without removing rotations.
Fix: Fix one person first. Use (n − 1)!.
Forgetting to divide by 2 for necklaces and garlands.
The word 'circular' triggers (n − 1)! automatically.
Fix: Read the object. If it can be turned over, use (n − 1)! ÷ 2.
Dividing by 2 for people at a table.
Students apply the necklace rule to all circles.
Fix: People cannot be flipped, so clockwise and anticlockwise are different. No division.
For 'together', using (n − 1)! × 2! instead of (n − 2)! × 2!.
They forget that the block counts as one object, reducing the total.
Fix: Count the units after grouping, say n − 1, then use (units − 1)! times the block's order.
Forgetting the gaps in an alternate arrangement.
Students multiply by the wrong factorial, ignoring that once one group of m people is seated, there are exactly m gaps for the second group.
Fix: Seat the first group in a circle in (m − 1)! ways, then arrange the second group in the m gaps in m! ways (when the groups are equal in size). If the second group has k people with k less than m, use mPk.
Applying the necklace division when beads are identical or n is 1 or 2.
The formula is memorised without its conditions.
Fix: Use the formula only for n ≥ 3 distinct objects. Otherwise count directly.

Worked examples

Example 1

In how many ways can 7 persons be seated around a round table? Options: (a) 5040 (b) 720 (c) 360 (d) 2520

Show the solution

The persons are distinct and cannot be flipped, so clockwise and anticlockwise are different.
Number of ways = (7 − 1)! = 6!.
6! = 720.

Answer: (b) 720

Example 2

How many different necklaces can be made using 6 distinct beads? Options: (a) 120 (b) 60 (c) 720 (d) 30

Show the solution

A necklace can be flipped, so clockwise and anticlockwise are the same.
Number of necklaces = (6 − 1)! ÷ 2 = 5! ÷ 2.
5! = 120, and 120 ÷ 2 = 60.

Answer: (b) 60

Example 3

In how many ways can 6 persons sit around a round table so that two particular persons A and B sit together? Options: (a) 120 (b) 48 (c) 240 (d) 24

Show the solution

Treat A and B as one block. Now there are 5 units to place around the table.
Circular arrangements of 5 units = (5 − 1)! = 4! = 24.
A and B can swap inside the block in 2! = 2 ways.
Total = 24 × 2 = 48.

Answer: (b) 48

Exam tips

Read the object first. Persons, seats and tables mean (n − 1)!. Beads, flowers, garlands and bracelets mean (n − 1)! ÷ 2.
Watch for n! or (n − 1)! among the options in a necklace question. These are common distractors that ignore the flip, so check them before you choose.
Watch for the word 'together' or 'alternate'. These problems are marked with blocks and gaps, and take more time, so attempt them after the direct ones.
Since a wrong answer costs 0.25 marks, attempt only when you have identified whether the arrangement can be flipped.

Practice questions from Permutations and Combinations

Circular Permutations: frequently asked questions

What is the formula for circular permutation?

For n distinct objects arranged around a circle, the number of arrangements is (n − 1)!. This applies when clockwise and anticlockwise orders are treated as different, as with people at a round table.

What is the difference between circular and linear permutation?

In a linear permutation, every order of n objects counts, so there are n! ways. In a circular permutation, rotations of the same arrangement count as one, so there are (n − 1)! ways.

Why do we divide by 2 for necklaces?

A necklace can be turned over, so an arrangement and its mirror image look the same. This halves the count to (n − 1)! ÷ 2. It applies to n ≥ 3 distinct objects.

How do I solve a round table problem where two people must sit together?

Join the two as one block, so you have n − 1 units. Arrange them in a circle in (n − 2)! ways. Then multiply by 2! for the order inside the block.