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  1. You are a host/hostess at your local Applebee’s. You are seating a group consisting of 9 couples at a round table.

    A)In how many different ways can you do this, provided that each couple will sit together, and all that you care about is their position relative to one another?

    B)What is the probability that Al doesn’t end up within two seats of Ricky, AND Beth doesn’t end up within two seats of Charlene?

Used a wrong tag before of order-statistics. Sorry about that.

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    $\begingroup$ What are your thoughts so far? It looks like you've copied a question without even all the necessary information $\endgroup$ – Ian Coley Nov 10 '13 at 21:24
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    $\begingroup$ This was all that was given to me by my instructor. I know that just using a simple 9 P 9 isn't enough, though. Because it's a circular table A,B,C,D,E,F,G,H,I would be the same thing as I,A,B,C,D,E,F,G,H. But then I get stuck. $\endgroup$ – Boxyouranswer Nov 10 '13 at 21:29
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    $\begingroup$ Depends. Al and Ricky could be a couple. $\endgroup$ – André Nicolas Nov 10 '13 at 21:30
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    $\begingroup$ Well, I assume from the question that they are not a couple. Otherwise it would just be 1/(the possibilities). $\endgroup$ – Boxyouranswer Nov 10 '13 at 22:39
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For the number of arrangements, imagine that one of the chairs is a throne, and the Queen is one of the group at Applebee's. She sits down first, of course, on the throne. Her Consort has $2$ choices of chair. Now let us seat the other couples one at a time, counterclockwise from the Queen-Consort pair.

The person chosen to occupy the chair immediately counterclockwise from the royal pair can be chosen in $16$ ways. Now the occupant of the next chair is determined. The person chosen to occupy the next chair after that can be chosen in $14$ ways, and then the occupant of the chair after that is determined. And so on.

Multiply. We get $(2)(16)(14)(12)\cdots (4)(2)$. This can also be written as $(2^9)(8!)$.

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  • $\begingroup$ Wow Thanks! But I have a follow-up question. Would the generic formula for such a problem then be 2^n * (n-1)! ? I mean, if there was another problem that said I have to seat 10 people around a round table with 10 available seats, could I use (2^10)(9!) ? $\endgroup$ – Boxyouranswer Nov 10 '13 at 22:52
  • $\begingroup$ Yes, precisely the same reasoning gives the expression in your comment, and works whenever we have $n$ couples. And if social arrangements divided people into triples, similar reasoning would give $(3!)^n (n-1)!$ ways to arrange $3n$ people around a round table so that "triples" are together. $\endgroup$ – André Nicolas Nov 11 '13 at 0:46
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Another approach to this counting problem is to think of each couple as "single" object, counting the number of ways the single objects can be arranged, and then multiplying by the number of equivalent ways the couples can be arranged as "compound" objects.

In particular, there are $9$ couples (considered as "single objects"), and they can be arranged around the table in $8!$ ways. (Make sure you understand why this is so)

Now we consider the couples as "compound" objects, consisting of persons $A$ and $B$. Each couple can be arranged in 2 ways: $AB$ or $BA$. Since there are $9$ couples, there are $2^9$ choices.

So we get $2^9 \cdot 8!$ arrangements.

And yes, this generalizes to arbitrary $n$: if there are $n$ couples, they can be arranged in $2^n (n-1)!$ ways.

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  • $\begingroup$ Follow-up. For part B of the problem, would I just find all the ways Al and Ricky do sit within 2 seats of each other, and all the ways that Beth and Charlene sit within 2 seats of each other, divide by the outcomes given by 2^9(8!) for each one then add together and subtract from 1? $\endgroup$ – Boxyouranswer Nov 10 '13 at 23:32
  • $\begingroup$ That is a valid approach. You can use the "compound" and "single" object approach, too. $\endgroup$ – nomen Nov 10 '13 at 23:37

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