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A total of 3n people, consisting of n families (father, mother and child), are randomly seated (all possible orderings being equally likely) at a round table.

Approximate the probability, for n large, that there are no families who are seated together.

I think that this is a poisson distribution problem since n is large. However, I am unsure how to approach this.. the answer should be exponential. I am already stuck at the first step of trying to find the probability of a family sitting together. I know that if this was a couple instead, the probability would be 2/(2n-1)! since the individual have 2 neighbours. In this case for a family, is it right that the probability is 3!/(3n-1)! since it can be seen as a permutation of 'F,M,C' If that is right, I am still unsure how to move forward from here. I know that my expectation would be 6n/(3n-1)!How do I apply the poison distribution here then?

choo
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Oct 11 '22 at 18:11
  • A possible intermediate step would be to scrutinize the Inclusion-Exclusion oriented analysis in this answer or the analysis to some of the other answers posted to the question. Using (for example) the inclusion exclusion framework of my answer, you would have to (somehow) take the limit, as $n \to \infty$ of each term. Personally, I don't know Poisson Theory from a giraffe. – user2661923 Oct 11 '22 at 18:19

1 Answers1

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Suppose one of the families is named Smith. We want to find the probability that the Smiths are seated together.

There are $(3n-1)!$ circular permutations of the $3n$ people, all of which we assume are equally likely. We want to count the number of those permutations in which the Smiths sit together. We assume $n > 1$. Pack the Smiths together and consider them as one object. Together with the remaining $3n-3$ people, there are then $3n-2$ objects to be arranged in a circle, which can be done in $(3n-3)!$ ways. The Smiths can be ordered in their packet in $3!$ ways, so the total number of arrangements with the Smiths sitting together is $3! \times (3n-3)!$. The probability that they sit together is then $$\frac{3! \times (3n-3)!}{(3n-1)!} = \frac{6}{(3n-1)(3n-2)}$$

Knowing the probability that the Smiths sit together, see if you can find the expected number of families sitting together. Then you might make the assumption that the number of families sitting together has a Poisson distribution in order to find the probability that none sit together.

awkward
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