Arrangements Around a Circular Table (Beyond 0606)

Not examined in 0606. Cambridge IGCSE Additional Mathematics 0606 (2025–2027) excludes permutations arranged in a circle. This page is extension practice, useful for A Level (9709) and SPM, not a 0606 exam question. For the P&C content that is examined, see the topic pillar.

Around a circular table there is no fixed first seat, so rotations count as the same arrangement. Fixing one person removes that repetition, giving $(n-1)!$ arrangements. A “must sit together” restriction uses the block method.

(a) In how many ways can $6$ people be seated around a circular table? (b) In how many of these do two particular people, $A$ and $B$, sit next to each other?

The working

Part (a), fix one person to remove rotations: $(6 - 1)! = 5! = 120$

Part (b), treat $A$ and $B$ as a single block. That leaves $5$ units (the $AB$ block plus the other $4$ people) around the circle: $(5 - 1)! = 4! = 24 \ \text{ways to seat the units}$

Within the block, $A$ and $B$ can swap: $\times\, 2! = 2$

So the number of arrangements with $A$ and $B$ together: $24 \times 2 = 48$

Where it goes wrong

• Use $(n-1)!$ for a circle, not $n!$. The "$-1$" is the whole point of circular counting.
• For the block, the number of units is $5$ (block $+ 4$), so it is $(5-1)! = 4!$, then $\times 2!$ for inside the block.
• Read whether seats are distinguishable (e.g. a numbered head of table). Here they are not, so the standard circular rule applies.

Common mistakes

• Using $6!$ for part (a).
• Forgetting the $\times 2!$ for $A$ and $B$ swapping inside the block.
• Counting $5!$ units in part (b) instead of $4!$ (using $n!$ not $(n-1)!$ for the reduced circle).

Topic home: Permutations & Combinations pillar. More: Worked examples.