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Eight people, including Abigail, Bethany, and Charlene, are to be seated at a circular table. Two seatings are considered distinct if, and only if, the ordering of people starting with Abigail and continuing clockwise around the table in one seating is distinct from that in the other seating. How many distinct seatings are there so that Bethany is between Abigail and Charlene?

The number of ways to seat the girls so that Abigail, Bethany, and Charlene are together is 5!. In a third of these arrangements - 40 of these arrangements - Bethany is between Abigail and Charlene.

Is this correct?

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There are $\frac{3!6!}{6}=3!5!$ ways in which Abigail, Bethany and Charlene are together. I think you tried to look at the three girls as one object, but you have yet to arrange them internally. With this in mind there are actually $5!\cdot2$ ways to arrange them so that Bethany is in the middle (since there are two internal arrangements of the three persons).

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  • $\begingroup$ Yes, I did consider the three girls together. So, that would almost be like seating 6 people around a table. There are 5! ways to do this. Oh, yes, and for each of these ways, either we would "take" either Abigail-Bethany-Charlene or Charlene-Bethany-Abigail in order. So, I should multiply this by 2. I think the answer is 2(5!). $\endgroup$ – user74973 Aug 23 '15 at 20:29
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    $\begingroup$ Yes, I agree with that. Nice approach by the way. $\endgroup$ – Jorge Fernández Hidalgo Aug 23 '15 at 20:31
  • $\begingroup$ Thanks for taking the time to comment. I have done probably a hundred such problems. Since I will be teaching on it, I wanted to discuss the solution with someone. $\endgroup$ – user74973 Aug 23 '15 at 20:33

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