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In how many ways can you seat the 12 people at 3 round tables such that:

A) All couples are seated together. (the two members of each couple sit side-by-side)

B) No couples sit together.

I've tried this question several ways, but keep getting different answers.

Since it is a round table problem, I've set the first seat arbitrarily and proceeded from there. Any help would be greatly appreciated.

Edit: For (A), I found that any two couples can be seated side-by-side 2 ways, thus I got 4*(4 choose 2) = 24 ways to seat the couples at two tables. Moreover, I found 6 ways of seating the 4 single people at the third table (by placing one person arbitrarily in the first seat, leaving 3 options for the second seat, 2 for the third, and 1 for the last).

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  • $\begingroup$ What do you mean by no couples sit together? $\endgroup$ Feb 6, 2019 at 20:30
  • $\begingroup$ The question was given ambiguously, but I'm assuming that one of the tables for (A) seats the 4 single people. $\endgroup$
    – user642272
    Feb 6, 2019 at 20:32
  • $\begingroup$ Have you tried using the principle of inclusion exclusion for (B)? $\endgroup$ Feb 6, 2019 at 20:40
  • $\begingroup$ I have, but I am unsure which sets I should use. I got 11! - 7!, which I believe is incorrect. $\endgroup$
    – user642272
    Feb 6, 2019 at 20:42
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    $\begingroup$ It seems like the problem was pretty clear that "all couples sit together" means that "for each couple, the two in that pair are next to each other." The only other reasonable interpretation of "all together" would be "all at the same table," which is impossible. $\endgroup$ Feb 6, 2019 at 20:51

1 Answer 1

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I assume the tables are interchangeable.

If you decide to seat two couples or a couple and two singles at a table there are four ways to seat them keeping the couple(s) together. If you seat four singles there are six ways.

For A you can either seat two couples at each of two tables and four singles at the third or seat two couples at one table and one couple and two singles at the other two. For the first there are three ways to split the couples, so $3*4*4*6=224$ ways to seat them. For the second there are three ways to pick the two couples, six ways to match two singles with the third, so $3*6*4*4*4=896$ ways to seat them. The total is $1120$

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