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There are four married couples sitting at a round table. How many ways are there to seat the eight people, if all couples should be sitting together?

My answer is $4! \cdot 2$ (4 couples and 2 variants of the shift, sitting on odd and even places).

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Is it correct?

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  • $\begingroup$ what is the meaning of 60" ? $\endgroup$ – miracle173 Jun 15 '17 at 11:58
  • $\begingroup$ It means the diameter of the table is 60 inches. $\endgroup$ – Especially Lime Jun 15 '17 at 12:21
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Le us name the couple aa,bb,cc,dd. The husband of coupl aa is a, the wife of aa is a. The members of the other couples are named similar. The places of the desk are numbered (clockwise) 1,2,3,...,8 (and next to 8 comes 1 again).

  • You have $8$ possibilities to reservate places for the first couple.

    • e.g. we place couple aa clockwise starting with seat 4:
    • the first couple is placed on the seats 4,5
  • You have $3!$ possibilities to specify the order of the couples following couple one in a clockwise sense.

    • after aa comes cc then bb then dd:
    • so cc has assigned 6,7
    • bb has assigned 8,1
    • dd has assigned 2,3
  • Then four each of the 4 couples you have 2 possibility to placethe husband and the wife on the seats assigned to the couple. So these are $2^4$ possibilities.

    • we choose:
    • Aa
    • Bb
    • cC
    • Dd

All in all you have $$8\cdot 3! \cdot 2^4=768$$ possibilities.

Our special selection is

1 2 3 4 5 6 7 8
b D d A a c C B

If one does not distinguish between arrangement that differ only by one of the 8 rotations then there are only

$$\frac{768}{8}=96$$

different positions.

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    $\begingroup$ The difference between this answer and the other two is that they assume two arrangements which differ by a rotation are the same (this is what's normally meant by a round table in these problems: there's no difference between the individual places so all that matters is where you sit relative to everyone else). $\endgroup$ – Especially Lime Jun 15 '17 at 12:20
  • $\begingroup$ As mentioned above, when you consider a round table you divide by the number of seats $n$. Also, $8 \cdot 3! \cdot 2^4 = 768$, and indeed, $\frac{768}{8} = 96$. $\endgroup$ – jvdhooft Jun 15 '17 at 12:21
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Treat each couple as a block.

You can order the couples around the table in $\displaystyle \frac{4!}{4}=6$ ways.

You can switch the order of the people in the couple in $2$ ways each, for $2^4$ ways.

The number of ways to seat the couples around the table is then, $6(2^4)=\boxed{96}$ ways.

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  • $\begingroup$ No, the right answer is 2*4! as it turned out. I'm trying to understand - why. $\endgroup$ – paus Jun 15 '17 at 9:21
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    $\begingroup$ @paus Given your original question, both our answers are correct. Can you quote the exact question from the textbook? $\endgroup$ – jvdhooft Jun 15 '17 at 9:26
  • $\begingroup$ Originally it's the same, but logically we have 4 couples to sit in two places, then we have three couples and so on until one couple. Then we can sit them either on odd or even places. I added the illustration to the main topic. $\endgroup$ – paus Jun 15 '17 at 9:41
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The number of ways to seat four people at a round table equals:

$$\frac{4!}{4} = 6$$

However, now they are couples, which can be seated in two unique ways. As such, the number of arrangements is:

$$\frac{4!}{4} \cdot 2^4 = 96$$

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