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I just need these checked. The symmetry in circular permutations is a bit confusing.

  1. What is the probability that 6 people sit in a circle in alphabetical order?
  2. How many ways can 6 people sit in a circle? Two arrangements are the same if you can rotate from one to the other.

For the first one, A can sit anywhere, and B has two options (either to the left or to the right of A). The order of the rest is then determined by this, so the probability is $$1 \cdot \frac25 \cdot \frac{1}{4!} = \frac{1}{60}.$$ For the second one, there are 720 ways of arranging them in a line, but for each arrangement, there are 6 equivalent ones (obtained by shifting). Hence the answer is 120.

user369210
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2 Answers2

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Yep, your work looks good to me.

Another way to do the first problem is to just count all the ways they can sit in order. There are $6$ choices for person $A$ to sit, and for each choice of seat there are only $2$ viable arrangements of the rest of them. Hence the probability is $\dfrac{12}{6!} = \dfrac{1}{60}$.

MT_
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(1) Yes; sound reasoning.   Alternatively, A' has six options, B' then has two options, and the rest are fixed by that; out of the $6!$ seating arrangements: $\dfrac{6\cdot 2}{6!}= \dfrac 1{60}$.

Note: the probability is half that if 'Alphabetic order' means ascending clockwise from A.

(2) Indeed.   There are $5!$ ways to seat $6$ in a circle when discounting (hexagon-)rotational symmetry.   I'm sure you can see the general pattern for seating $n$ in a circle.

Graham Kemp
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