# Seating arrangements at a circular table

There are 6 women and 6 men to seat at a round table. Suppose that only the pattern below must be taken into account:

'H' should mean 'M' and 'M' should mean 'W'

How many ways can one seat them at a round table following the restriction explained above?

I thought of fixing the first man and then consider $$\mathrm{ 5!}$$ choices for the remaining ones and after them being chosen, I can consider $$\mathrm{ 6!}$$ choices for women, so the final answer would be $$\mathrm{ 6!*5!}$$ but is there something left to be considered?

• What does "H" mean? If "M" spots are required to be men, '"H" can only mean women but still should be clarified. Sep 26, 2023 at 16:36
• Does your first man have a woman on his right, or on his left? Sep 26, 2023 at 16:36
• Both cases should be considered
– J P
Sep 26, 2023 at 16:42
• What do you mean by "first man" ? Every man is distinct, and on an unnumbered table, no one is "first" Sep 26, 2023 at 17:05
• We can in problems like this "assume without loss of generality that PersonA sits at the northernmost seat at the table" and have people sit around them, using that person as our reference point. Sep 26, 2023 at 17:06

Your approach is good however it does miss one nuance which is pointed out in the comments.

We can assume without loss of generality that our people are named Man1, Man2, ..., Man6 as well as Woman1, Woman2, ..., Woman6.

We can assume without loss of generality that Woman1 sits at the northernmost seat at the table.

We can then arrange everyone else around the table relative to this first person seated. There are two cases two consider... Woman1 has a woman to her right and man to her left, or vice versa. Once that decision has been made, all seats will unambiguously be reserved for a specific gender. Decide which seat designated for men gets occupied by each remaining man, and similarly for the women.

This gives us a total of $$2\times 6!\times 5!$$ which one will notice is equal to $$\dfrac{6!\cdot 6!}{3}$$ and not equal to $$\dfrac{6!\cdot 6!}{12}$$. The punchline here is that "division by symmetry" arguments can be flawed, especially when falling into habits thinking things like "circular tables means divide by size of table" which is not necessarily true.

This can also be described with language of (the lemma which is not) Burnside's Lemma. Here, we have three group actions able to act on this that keep the pattern the same. Rotating by four spaces, rotating by eight spaces, or not rotating at all, which explains the division by three.

• Could you clarify in this solution - are "rotations" of a given seating considered the same, so it is literally the possible orders of the 12 people being counted? Sep 26, 2023 at 19:04
• @AlgTop1854 yes, as implied in the problem by mentioning that the table is "circular." The rotating I refer to in my last paragraph was specifically so that the M's and H's appear not to have moved... that I would consider MMHHMMHH... different than MHHMMHHM... and so on, this again to properly use (not) burnside's lemma Sep 27, 2023 at 12:48
• Thank you and it's clearer now! My first thought was just $6!6!$ which missed this and would be the answer if the question was just seating 6M and 6W in a row of seats numbered 1 to 12. The point about left or right of the first M still seems confusing because (assuming M is a man) the picture implies that the M at the top has an H on his right and an M on his left. Sep 27, 2023 at 16:32
• @AlgTop1854 if you preferred to ensure that the labeling is accurate to the picture... you could go through the same argument as I have above... however begin not by allowing Woman1 to sit at the northernmost seat WLOG... instead say that Woman1 sits in one of the two northernmost seats (designated for women) and pick which one. This... instead of orienting ourselves with respect to Woman1's seat... we instead orient ourselves with respect to the pair of seats for women that woman1 is occupying one of. Sep 27, 2023 at 16:39