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How many ways to seat 4 couple and 2 single around a round table, provided that each couple will sit together

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  • $\begingroup$ if seats have number, or seats dont be same what happend? $\endgroup$ – DnShVr Sep 30 '14 at 16:59
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Edit: Sorry for the confusion!

Viewing each couple as a unit, you're arranging six units around a round table, for which there are $5! = 120$ ways, since we can fix one of the singles to a chair. Then, you can change the order within each couple, so there should be $120 \times 2^4 = 1920$ ways.

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  • $\begingroup$ but why we set 6!, six units around a round table, We dont should take 5!? rotation of units are considered!! $\endgroup$ – DnShVr Sep 30 '14 at 17:11
  • $\begingroup$ @DnShVr Sorry for the confusion; I fixed it, and it's my final answer. $\endgroup$ – symmetricuser Sep 30 '14 at 17:32
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    $\begingroup$ Worth pointing out that since symmetricuser's edit, paw's comment no longer applies. $\endgroup$ – gj255 Sep 30 '14 at 18:22
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In "round table" problems, the convention is that two arrangements that differ by a rotation are to be considered "the same." Equivalently, if we are talking about arranging distinct objects, we can fix arbitrarily the position of one object.

Imagine that one of the singles is the Queen, and one of the chairs is a throne. Of course, the Queen sits on the throne. Now we wish to arrange the rest of the people.

We have $4$ couples left, and $1$ single. These $5$ objects can be arranged in $5!$ ways. For every such way, the person with the lower student number in the couple can sit immediately to the left or immediately to the right of the person with the higher student number. The total number of seatings is therefore $5!2^4$.

Remark: The number $5!2^4$ was the first answer of symmetricuser.

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  • $\begingroup$ Ah thank you for clarifying... I was confusing myself after thinking why I should divide by 10. $\endgroup$ – symmetricuser Sep 30 '14 at 17:30
  • $\begingroup$ True to your name, you were trying for full symmetry. However, the setup is not symmetrical, because of the single/couple division. $\endgroup$ – André Nicolas Sep 30 '14 at 17:35

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