The question is:

-How many ways are there for $n$ people to sit around a circular table if two seating arrangements are considered identical if each person has the same left neighbor in them?

I recall from my probability course long ago that for circular arrangements the formula is $(n-1)!$ but that doesn't seem correct to me. Am I missing something in my answer?


$(n-1)!$ is correct.

You start by placing the first person anywhere in the table. Any starting placement is identical because of the circular symmetry.

Then you have $(n-1)$ choices for the person sitting to their left. Then $(n-2)$ choices for the person sitting to the left of the second person and so on. Thus $(n-1)!$ choices overall.

It would help if you explained why you think the formula was wrong so we can address your particular concerns.

  • $\begingroup$ Thank you! The way the question was worded, that second part, confused me for some reason. I do understand it now though. Great explanation btw! $\endgroup$ – user482578 Oct 4 '17 at 16:33

Why wouldn't it be correct?

Take a particular person and count distinct arrangements of the $n-1$ people to hir left (clockwise around the table, wherever zhe may sit).   Are there not $(n-1)(n-2)\cdots 1$ ways to so choose left-person successors?

Well, there you go.   That is $(n-1)!$



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