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The Question

How many can we arrange $3$ men and $3$ women around a circular table in such a way that no man sits with another man, and no woman sits with another woman. Combinations that can be obtained from another seating arrangement by rotation are considered identical. This was given as a problem by my professor at the end of lecture. Already read this question I got a different answer than their accepted answer, which I don't think is right.

My Work

The first step is to determine how many ways to arrange men and women alternating in a line with no restrictions. First we must decide if we are starting with a man or a woman, there are $2$ choices for this. From there, regardless of choice there are $3*3*2*2*1*1$ ways to arrange the ladies and gents. Now we account for all the rotations there are 6 rotations, so we divide by 6 to avoid overcounting. $\frac{2*3*3*2*2*1*1}{6} = 3*2*2*1*1 =12 $ ways to arrange them like this.

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Dunka
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  • you are multiplying by $2$ for the decision whether to start with a man or a woman. but if you start with a man, and then rotate you get the same arrangement starting with a woman which you want to avoid overcounting. so shouldn't you not multiply by $2$? – benji Jan 09 '15 at 22:48
  • My logic was that say we have a head chair where we will place our first guest, it'll make a difference if the person at the head chair is a female instead of a male, so I counted it in my initial calculation. – Dunka Jan 09 '15 at 22:52

1 Answers1

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To avoid the initial overcounting in the knowledge that rotations will be ignored, you can arbitrarily reserve one seat to a particular woman. Then there are two ways to seat the other two women, and 3!=6 ways to seat the men.

As you say, 12 total seatings ignoring rotations.

As a further refinement, you could also ignore reflections - that is, if each person has the same neighbours in two patterns, the seating patterns are regarded as equivalent. This would lead to only 6 possibilities.

Joffan
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