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Suppose you have 12 people sitting around a round table, what is the probability that A,B,C and D sit evenly spaced around the table?

The total number of ways is $11!$ and the number of sitting evenly spaced is, I think, $4!8!$ because there are $4!$ ways to sit them leaving two seats between them and there are $8!$ ways of filling those sits with the rest of the people; so the probability is $\frac{4!8!}{11!}$. Is this correct?

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I don't believe you are accounting for the rotational symmetry when counting the number of ways to evenly space them. For example, suppose we always seat $A$ so they are at the top of the table, then as you pointed out there are $11!$ ways to seat the other people around the table. But then there are 3! ways to seat the other 3 people so all 4 are evenly spaced, and 8! ways to seat the rest in between.

I believe this makes the probability $\frac{3!8!}{11!}$

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Your numerator should be $3! 8!$. You can seat person A arbitrarily. There are then $3$ seats for B, C, D; and $8$ seats for the others.

Or, alternatively, there are ${12} \choose 4$ sets of $4$ people. Three of those sets will be spaced evenly. So there should be a probability of $\frac{3}{{12} \choose 4}$ of getting your specific group of four spaced evenly.

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  • $\begingroup$ I don't believe $\frac{3}{\binom{8}{4}}$ is equal to $\frac{3!8!}{11!}$. Can you expand on your second solution? $\endgroup$ – ClownInTheMoon Sep 30 '16 at 3:11
  • $\begingroup$ @ClownInTheMoon My mistake. Those should be ${{12}\choose 4}$. I'll fix it. $\endgroup$ – paw88789 Sep 30 '16 at 6:15

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