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Eight friends sit round a table. A linear-arrangement habit says 8! = 40,320 ways, and the answer key says 5,040. The difference is not a formula to memorise. It comes from asking one question every time: which arrangements look the same? This post answers it for tables, numbered seats and necklaces, with worked problems and the slips that cost marks.
Where is this in the syllabus?
The NCERT Class 11 chapter Permutations and Combinations (2026-27 reprint) builds everything from the fundamental principle of counting in section 6.2. It has no separate section on circular arrangements, so the method below is an application of that principle. The JEE (Main) 2026 syllabus lists permutations and combinations "and simple applications" in Unit 4, and the JEE Advanced 2026 brochure includes permutations and combinations and "computation of probability of events using permutations and combinations". Check the syllabus for your exam year.
Why is the answer (n − 1)! and not n!?
Seat n different people in a row of n chairs: n! ways. Now bend the row into a circle. Rotating everyone one seat to the left gives a seating where each person has the same neighbours on the same sides. Every circular seating appears n times among the n! rows, once for each rotation. So the count is n!/n = (n − 1)!.
A quicker way to see it: fix one person's seat. That kills the rotations. The other n − 1 people then fill the remaining seats in (n − 1)! ways.
When do you divide by 2?
People at a table face inwards, so clockwise and anticlockwise orders are different: A's left neighbour becomes A's right neighbour. A necklace or garland has no inside. You can flip it over, which turns a clockwise order into an anticlockwise one. So for n different beads, with n at least 3, the count is (n − 1)!/2.
And when seats are numbered or one chair is marked, rotations no longer look the same, so the count goes back to n!.
Worked example 1: together and apart
Eight people, including A and B, sit round a table.
No condition: 7! = 5,040.
A and B together: tie them into one block. Seven units round a table give 6! = 720 ways, and the block can be AB or BA, so 720 × 2 = 1,440.
A and B apart: total minus together, 5,040 − 1,440 = 3,600.
Worked example 2: alternate seating
Four men and four women sit round a table so that no two men are next to each other.
Seat the men first: 4 people round a table give 3! = 6 ways. That leaves exactly four gaps between them, and the four women fill those in 4! = 24 ways. Total: 6 × 24 = 144. Do not use (4 − 1)! again for the women. Once the men are seated, the gaps are fixed positions.
Worked example 3: choose, then seat
From 10 people, choose 6 and seat them round a table. Choose first: C(10, 6) = 210. Seat them: 5! = 120. Total 210 × 120 = 25,200.
Check another way: P(10, 6) = 151,200 ordered rows, and each circle appears 6 times, so 151,200/6 = 25,200.
Worked example 4: a garland
Ten different flowers make a garland. Flips count as the same, so the answer is 9!/2 = 181,440.
Worked example 5: a probability twist
Eight people sit at random round a table. What is the probability that A and B sit together? From example 1, 1,440/5,040 = 2/7.
Check: fix A's seat. B is equally likely to be in any of the other 7 seats, and 2 of them are next to A. So 2/7 again. When two methods agree, you can trust the count.
Which mistakes cost marks?
Halving for people at a table. Only flippable objects get the /2.
Forgetting the internal order of a block, the × 2 in example 1.
Using (n − 1)! twice in alternate seating.
Dividing when seats are numbered.
Using (n − 1)!/2 with identical beads. The formula assumes every bead is different. With repeated colours, some arrangements map onto themselves when rotated, so simple division breaks. List small cases instead.
For more on avoiding double counting in general, read our guide to permutations and combinations.
When you cannot tell whether two arrangements are the same, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer, which helps you see where each division comes from.
A practice routine
For every circle problem, write one line first: "can rotate", "can rotate and flip" or "fixed seats".
Solve each problem two ways where you can, as in examples 3 and 5.
Check one small case by listing: 3 people round a table should give 2 seatings.
Mix circle problems with linear ones so you practise deciding, not just applying.
Try this now: in how many ways can 5 boys and 5 girls sit round a table so that no two girls are together? (Answer: 4! × 5! = 2,880.)
