A probability problem: In how many different ways can 5 people sit around a round table? Is the symmetry of the table important?

Answer: If the symmetry of the table is not taken into account the number of possibilities is 5! = 120. In this case it would be the same as ordering people on a line. However if rotation symmetry is taken into account, there are five ways for people to sit at the table which are just rotations of each other. So using symmetry the answer is 24.

So I understand why it's 5! for the first part. Order matters so it's a permutation without replacement, so $\frac{n!}{(n-r)!}$ = $\frac{5!}{(5-5)!}$ = 5! = 120

But what about the last part, is there any way to describe that as a permutation/combination? How exactly is 24 calculated? (it appears to be $\frac{5!}{5}$, but I don't see why)

Thank you in advance for any explanations!

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    – Tavrock
    Mar 2, 2017 at 6:45

1 Answer 1


If you count it as you initially did, $5!$, you are over counting. For every arrangement, say, (May, Paul, Tina, Fran, Bob), there are $5$ arrangements counted within $5!$ that are inditinguishable in the circular setting. For instance, (Paul, Tina, Fran, Bob, May) or (Tina, Fran, Bob, May, Paul), (Fran, Bob, May, Paul, Tina) and (Bob, May, Paul, Tina, Fran) all considered identical to the initial (May, Paul, Tina, Fran, Bob). These are cyclic permutations.

They are counted as $(n-1)!$ Instead of $n!$

This is the same as in your case adjusting the overcounted $5!$, by dividing by $5$:


The premise is that you are simply rotating the face of a clock, leaving the relative positions unchanged. There is no external reference, only who is sitting to the left and to the right of each person.

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    $\begingroup$ That's assuming (and I don't really doubt that you correctly guess the original intent) that being in the seat nearest the window rather than the seat nearest the door makes no difference -- if it did, those 5 arrangements become distinguishable. $\endgroup$
    – Glen_b
    Mar 2, 2017 at 9:30
  • $\begingroup$ @Glen_b This is what I tried to convey in the last sentence of the last paragraph. $\endgroup$ Mar 2, 2017 at 11:24
  • $\begingroup$ Yes, you did make it clear that was the assumption there; I guess the question kind of covers the "a position can be distinguished" case $\endgroup$
    – Glen_b
    Mar 2, 2017 at 11:53

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