Count arrangements of n distinct objects around a circle as (n-1)! by treating rotations as identical, extend this to groups and blocks around a circle, alternating patterns and 'not together' restrictions handled by the complement, and recognise that for a necklace or bracelet, where a reflection also coincides, the count is (n-1)!/2

What this dot point is asking

The multiplication principle and ordered selections page counted arrangements in a line: $n$ distinct objects in a row give $n!$ orderings, because the first position has $n$ choices, the next $n-1$, and so on. This page changes the geometry. When objects are placed around a circle, a round table, a ring of beads, numbers on a clock face, there is no first seat and no last seat, just a cyclic order of who sits next to whom. The new question is which arrangements should count as the same, and the answer reshapes the count from $n!$ down to $(n-1)!$.

The single idea driving everything here is that a rotation does not make a new arrangement. If everyone at a round table shuffles one seat clockwise, the seating "looks" different in a photo but nobody's neighbours have changed, so it is the same circular arrangement. Because each circular arrangement can be rotated into $n$ different line-ups (one for each starting seat), the $n!$ line arrangements count every circular arrangement exactly $n$ times, and dividing gives $\dfrac{n!}{n} = (n-1)!$. From that one principle the rest of the dot point follows: arranging groups or blocks around a circle (couples who must sit together), forcing an alternating pattern (boys and girls), keeping a pair apart by counting the complement, and the special case of a necklace or bracelet, where the ring can also be turned over, so a reflection coincides too and you divide by a further $2$ to get $\dfrac{(n-1)!}{2}$.

The answer

Why $n$ distinct objects around a circle give $(n-1)!$

Start with a concrete round table. King Arthur, Guinevere, Lancelot, Bors and Percival sit around it. As a line of five seats there would be $5! = 120$ orderings. But around a circle, the five seatings you get by rotating one fixed order, everyone moves one seat clockwise, then another, and so on, are all the same arrangement: the cyclic order, and therefore everyone's left and right neighbour, is unchanged.

The figure shows four of the five rotations of one seating across the top; all are the same arrangement, so they funnel into the single circular seating at the bottom. Because every distinct circular seating produces exactly $5$ of the $120$ line arrangements, the number of genuinely different circular seatings is

The same argument works for any $n$: there are $n!$ line arrangements, each circular arrangement appears $n$ times among them (one per starting position), so the circular count is $\dfrac{n!}{n} = (n-1)!$.

The fastest way to see it: fix one object

There is a quicker route to the same $(n-1)!$ that is also the cleanest method in the exam. Since rotating changes nothing, you may fix one object in one seat and lose no generality, this single choice "anchors" the circle and uses up all the rotational freedom at once. Once one person is seated, the remaining seats are now genuinely distinguishable (each is a definite distance clockwise from the anchor), so the other $n-1$ objects fill them like an ordinary line.

With one person anchored at the top, the remaining four seats are filled in $4 \times 3 \times 2 \times 1 = 4! = 24$ ways, the same $(5-1)!$ as before. This "fix one, then count the rest as a line" move is the workhorse for every restricted circular problem below: deal with any special object by anchoring or blocking first, then arrange whatever is left.

Groups and blocks around a circle

When some objects must stay together, use the same block method as for lines, adapted to the circle. Glue each group into a single block, arrange the blocks around the circle as $(\text{blocks} - 1)!$, then multiply by the number of ways to order the members inside each block. The only circular twist is that arranging the blocks is itself a circular arrangement, so it is $(\text{blocks}-1)!$, not $(\text{blocks})!$.

The cleanest example is couples. Suppose $k$ couples sit around a table with each couple together. Treat each couple as a block: there are $k$ blocks, arranged around the circle in $(k-1)!$ ways, and each couple can sit in $2$ internal orders, giving $2^k$ across the $k$ couples. The total is $2^k \times (k-1)!$.

For $3$ couples the figure gives $2^3 \times (3-1)! = 8 \times 2 = 16$, and a rotation-aware enumeration of the $6$ people with every couple adjacent confirms $16$. The same recipe scales: $5$ couples give $2^5 \times (5-1)! = 32 \times 24 = 768$. If instead all the boys sit together and all the girls sit together (two blocks of $5$), you get $5! \times 5! \times (2-1)! = 120 \times 120 \times 1 = 14\,400$, the two blocks arranged around the circle in $(2-1)! = 1$ way, with each block internally ordered in $5!$.

Alternating patterns

A favourite restriction is alternating two types around a circle, boys and girls, odds and evens, two colours, in strict $\text{ABAB}\ldots$ order. With $n$ objects of each type, the clean method is: seat one type first as a circle, then slot the other type into the gaps.

Seat the $n$ objects of the first type around the table in $(n-1)!$ ways (this absorbs the rotations). Their placement creates $n$ gaps between them, and because the rotations are already used up, those gaps are now distinct seats; the $n$ objects of the second type fill them in $n!$ ways. So the count is

For $3$ boys and $3$ girls this is $(3-1)! \times 3! = 2 \times 6 = 12$, and a rotation-aware enumeration confirms $12$. For $5$ of each it is $(5-1)! \times 5! = 24 \times 120 = 2880$. The asymmetry, $(n-1)!$ for the first group but $n!$ for the second, catches many students out; it is exactly because seating the first group is the circular step that fixes the rotations, after which the gaps are ordinary distinct seats.

"Not together": count the complement

When a pair must sit apart, counting the seatings where they are apart directly is fiddly. The reliable move is the complement: count the total, count the (easier) "together" case with the block method, and subtract.

Take $n$ people around a table with a particular pair $A$ and $B$ who must not be adjacent. The total is $(n-1)!$. The "$A$ and $B$ together" seatings glue them into a block: that is $n-1$ units around the circle, $(n-2)!$ arrangements, times $2$ for the internal order of the pair, so $2 \times (n-2)!$. Therefore

For $n = 6$ this is $5! - 2 \times 4! = 120 - 48 = 72$, and a rotation-aware enumeration confirms $72$ seatings with $A$ and $B$ apart (and $48$ with them together, totalling $120$). For $n = 5$ it is $4! - 2 \times 3! = 24 - 12 = 12$. The complement also feeds probability directly: if the six sit at random, $P(A,B\text{ apart}) = \dfrac{72}{120} = \dfrac{3}{5}$.

Necklaces and bracelets: divide by 2 for reflection

A round table has a fixed top and bottom, so two seatings that are mirror images are genuinely different (your left-hand neighbour is not your right-hand neighbour). A necklace, bracelet or key-ring is different: you can pick it up and turn it over. Flipping it reverses the cyclic order, so an arrangement and its mirror image are the same object. On top of identifying rotations, you must now also identify each arrangement with its reflection.

The figure shows a ring of $5$ beads and its mirror image (the same beads read the other way round); turning the necklace over carries one into the other, so they count once. Starting from the $(n-1)!$ round-table count, the reflection pairs the arrangements into mirror twins, so you divide by a further $2$:

For $5$ distinct beads this is $\dfrac{4!}{2} = \dfrac{24}{2} = 12$, and a direct enumeration that identifies both rotations and reflections confirms $12$. For $6$ distinct beads it is $\dfrac{5!}{2} = \dfrac{120}{2} = 60$; for $10$ different keys on a key-ring, $\dfrac{9!}{2} = \dfrac{362880}{2} = 181\,440$. The "divide by $2$" is clean only when no arrangement is its own mirror image, which holds whenever the beads are all distinct and $n \ge 3$.

Using circular counts in probability

Once you can count circular arrangements, probabilities follow from $P = \dfrac{\text{favourable}}{\text{total}}$, with both counts done by the methods above. For example, three Tasmanians, three New Zealanders and three Queenslanders ($9$ people) sit at random around a round table; the chance the three state-groups each sit together is

The numerator treats each group as a block: order inside each group in $3!$ ways ($(3!)^3$ across the three groups), then arrange the $3$ blocks around the circle in $(3-1)! = 2$ ways. The denominator is the unrestricted $(9-1)! = 8!$. A smaller analogue, $2$ from each state ($6$ people), gives favourable $(2!)^3 \times (3-1)! = 8 \times 2 = 16$ out of $(6-1)! = 120$, and a rotation-aware enumeration of that analogue confirms the $16$.

Verifying a small case by listing

Whenever a circular count is small, you can list the genuinely different arrangements to check the formula, the discipline being to fix one object so you never write a rotation twice. Take $4$ people $A, B, C, D$ around a table. The formula gives $(4-1)! = 3! = 6$. Anchoring $A$ at the top and reading clockwise, the distinct seatings are

which is $6$ arrangements, matching $(4-1)!$. Fixing $A$ guarantees each circular seating is written exactly once (its rotations all start with $A$ when read from the anchor). If instead this were a necklace, $ABCD$ and its reverse $ADCB$ would be the same, pairing the six into three, $\dfrac{(4-1)!}{2} = 3$, which a reflection-aware enumeration also confirms.

How exam questions ask about this dot point

The wording tells you which move to reach for. Map the phrase to the method:

• "...around a (round) table / in a circle / in a ring", "seated around". Plain circular arrangement: $(n-1)!$, because rotations are the same.
• "...how many ways in a line / in a row" then "...in a circle". The line is $n!$; the circle is $(n-1)!$, a factor of $n$ smaller.
• "...must sit together / as a group / a particular couple together". Block method: glue the group, arrange the blocks as a circle $(b-1)!$, times the internal orders. Couples: $2^k \times (k-1)!$.
• "...boys and girls alternate / alternating colours / odds and evens alternate". Seat one type as a circle then fill the gaps: $(n-1)! \times n!$ (needs equal numbers).
• "...must not sit together / sit apart / separated / not adjacent". Complement: total minus together, $(n-1)! - 2 \times (n-2)!$ for one pair.
• "...immediately to the left/right of", "next to / between two named people". Fix the relationship as a forced (or two-way) block, then arrange the remaining units around the circle.
• "...a necklace / bracelet / key-ring", "the ring can be turned over", "reflections are the same". Divide the round-table count by a further $2$: $\dfrac{(n-1)!}{2}$.
• "...probability that ... around the table". $P = \dfrac{\text{favourable circular count}}{(n-1)!}$, both counts by the moves above.