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I'd like to know if my solution is plausible.

4 couples are to be seated around a circular table. How many ways can they be seated if each person is sitting directly across from their spouse (i.e. there are three people between them and their spouse on either side)?

I went like this. Having to a round table gives me 7!2! to sit the first couple. For the second couple I have 6!2! to sit them (since I have 6 spot for the first person his spouse and him can just sit across which make for 2!) I managed to continue with that way of thinking to get in final.

7!2!6!2!4!2!2!

is this answer plausible ? thank you and have a nice day

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  • $\begingroup$ Please delete your accepting my answer. My answer is wrong!!! $\endgroup$ – zoli Feb 3 '18 at 12:24
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Unless otherwise specified, in circular seating arrangements, only the relative order of the people matters. Therefore, we may consider seating arrangements relative to the first person we sit.

Line up the women up in alphabetical order. Seat the first woman. Her husband must sit across from her. The next woman can be seated in one of six remaining seats. Her husband must sit across from her. That leaves four seats for the third woman. Her husband must sit across from her. The final woman can be seated in one of the two remaining seats. Her husband must sit across from her. Hence, there are $6 \cdot 4 \cdot 2 = 48$ distinguishable seating arrangements in which each husband and wife sit across from each other.

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  • $\begingroup$ I don’t get why you would fix the first couple and then sit everybody around them. Also if I was given just 5 persons to sit around the table I would get them sorted in 4! ways right ? $\endgroup$ – Mahamad A. Kanouté Feb 2 '18 at 21:28
  • $\begingroup$ It is true that five people can be seated in $4!$ ways. We could seat one of those five people at the table, then seat the remaining four people in $4!$ orders as we move clockwise around the table from the person we seated first. In this case, we seat one of the woman. Since her husband must sit across from her, he can only be seated in one way. That leaves six seats for the next woman to choose. Once she makes her choice, her husband must sit across from her. That leaves four seats for the third woman to choose. Her husband must sit across from her. The last woman has two choices. $\endgroup$ – N. F. Taussig Feb 2 '18 at 21:41
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Here's a different approach. There are four surnames A B C and D. If you draw a dividing line across the middle of the table, then on one side you need to arrange the letters A B C D which can be done in $4!$ ways.

For each of these surnames, there is a husband or a wife who occupies this position, with their spouse automatically sat in the place opposite. We can choose whether each surname is represented by husband or wife in $2^4$ ways.

Finally, for each arrangement, we can ask everyone to move one place to the left, but we still have the same arrangement. Likewise we can rotate each arrangement into eight new positions but each arrangement is still the same as the original. We therefore divide by eight.

Putting this all together we have a total of $$\frac{24\times2^4}{8}=48$$ arrangements.

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