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I'm practicing for the actuarial examination and I found this problem that I couldn't solve. If someone can help me, I will be happy :-).

Five Americans, three Germans and four Italians go to dinner together. They randomly sit at a round table with twelve chairs. What is the probability that the people of the same nationality sit together?

5----American
3----German
4----Italian

Total=12  Chairs=12   Total outcomes= 12!

Then, I was thinking about sitting first the Americans together and then the Germans and then the Italians together, but there are so many possibilities that I believe can't no be right. Are there any quick way? or correct one?

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DarioC
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    At least one pair of one nationality together or everyone from each nationality together? – Henry Dec 13 '15 at 19:41
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    The total number of outcomes is $(12 - 1)! = 11!$ since the relative positions of the people are left unchanged by the $12$ rotations of the people. – N. F. Taussig Dec 13 '15 at 19:54

1 Answers1

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Seating in a circle, only the relative positions matter. First consider only blocks of each nation.

Seat the Americans together anywhere. The Germans must either be to their left or to their right, and then the Italian block gets fixed automatically.

Thus $2$ ways for blocks to be arranged, and $5!3!4!$ ways for people to be permuted within the blocks, hence 2!5!3!4! distinct arrangements.

Oh, and since you are asking the probability, $Pr = \dfrac{2!5!3!4!}{11!}$

[ Note that for unnumbered seating in a circle, one position is "lost" in creating a reference point for all the others, hence 11! unconstrained seating arrangements, not 12! ]

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