A round table has no first seat — spinning it does not reseat anyone.
The problem
In how many distinct ways can 5 people be seated around a circular table?
The answer is NOT 5 factorial.
Pin one person down to kill the rotations, then arrange the other four freely: (5 − 1)! = 4! = 24. D is 5! = 120, which counts every rotation of the same seating as a new one — around a table there is no first chair.
Everyone still has the same neighbours.
All three are the SAME seating
ABCDE and BCDEA are two different line-ups, but round a table they are one arrangement seen from a different chair.
Pretend the seats are numbered, then fix it afterwards.
seat 1
5
×
seat 2
4
×
seat 3
3
×
seat 4
2
×
seat 5
1
5! = 5 × 4 × 3 × 2 × 1
one per seat
= 120
the line-up count
But each real seating got counted several times.
Five chairs means five spins that all look the same.
spin 0
ABCDE
spin 1
BCDEA
spin 2
CDEAB
spin 3
DEABC
spin 4
EABCD
5 copies of ONE seating
120 ÷ 5 = 24
throw away the duplicates
n people means n spins, so you always divide by n.
circular = n!n = (n − 1)!
One n cancels off the front of the factorial.
| People | Line-ups | Around a table |
|---|---|---|
| 3 | 3! = 6 | 2! = 2 |
| 4 | 4! = 24 | 3! = 6 |
| 5 | 5! = 120 | 4! = 24 |
seat 1, seat 2, …
use 5! = 120
spins now DO differ
you can flip it over
use (n − 1)!2
flips count as the same
Our table is plain and unlabelled, so 24 stands.
Numbered seats with spare chairs are an ordinary nPr: 3 people in 5 numbered chairs can sit in 5 × 4 × 3 = 60 ways. Each person picks a different labelled chair, and no division is needed.
Can you tell the seats apart without looking at the people? If yes, it is an ordinary permutation.
glue the friends: 6 units around the table
the pair is one unit
(6 − 1)! = 120
the circle rule on the units
120 × 2 = 240
the friends can swap
“Not together” is everything minus “together”:
all seatings: (7 − 1)! = 720
720 − 240 = 480
not together
The units still sit round a table, so it is (6 − 1)!, not 6!. 6! × 2 = 1440 counts every rotation separately.
seat the men: (3 − 1)! = 2
the circle rule, used once
the 3 gaps between the men are now different seats
the rotation is already used up
the women fill the gaps: 3! = 6
an ordinary line count
2 × 6 = 12
(3 − 1)! × (3 − 1)! = 4 lets the women’s seats rotate again; 3! × 3! = 36 forgets the table is round. Only the first group uses the circle rule.
Only some sit. In how many ways can 4 of 7 people be seated around a round table?
choose who sits: 7C4 = 35
a group, no order yet
seat them: (4 − 1)! = 6
the circle rule on the 4
35 × 6 = 210
same as 7P4 ÷ 4 = 840 ÷ 4
What is different: the circle rule uses the number who sit, not the number in the room. Choose first, then go round.
(7 − 1)! = 720 seats all seven. With only four chairs, the rotations are spins of four seats, so divide 7P4 by 4.
One person between two others. Six people sit around a round table. In how many ways can they sit if A sits between B and C?
block: B A C or C A B → 2 inside orders
A is fixed in the middle
units: the block and 3 others = 4 → (4 − 1)! = 6
around the table
6 × 2 = 12
both stages
What is different: the block has a fixed middle, so it has only 2 inside orders, not 3! = 6.
Directly opposite. Six people sit around a round table. In how many ways can they sit if A and B must face each other?
seat A anywhere: 1 way
pinning A uses up the rotation
B takes the seat across: 1 way
only one seat is opposite A
the other 4 fill 4 labelled seats: 4! = 24
an ordinary line count
What is different: once A is pinned, every other seat is fixed relative to A, so the rest is a plain factorial with no extra division.
4 couple-blocks around the table: (4 − 1)! = 6
each couple can swap: 2 × 2 × 2 × 2 = 16
one factor of 2 per couple
6 × 16 = 96
Each couple swaps on its own, so the factor is 2⁴, not 2. Using 2 gives 12.
Round-table items almost always add a condition: two people together or apart, groups alternating, couples side by side, or numbered seats that cancel the circle rule. Use (n − 1)! once — on the units or on the first group — and ordinary factorials for everything else. Ten-second check: “together” plus “apart” must add up to the plain (n − 1)!.
The other four can sit in any order — 4! = 24
The problem
In how many distinct ways can 6 different keys be arranged on a key ring?
A ring is circular, so start with (6 − 1)! = 120 — but a key ring can also be turned over, which makes each arrangement identical to its mirror image. Divide by 2: 60. A forgets the flip, and it is the flip that separates a ring from a table.
Hint: a key ring can be turned over as well as spun, so there is one more division to do after (n − 1)!.
