This is in connection with my previous question. Suppose the question is "How many ways can three married couples sit around a round table if husband and wife must sit opposite each other"

My approach is given below. One husband can be seated in a seat which is indistinguishable and his wife at an opposite seat. then the positions are identifiable and remaining persons can be placed in 4*2 = 8 ways. total 8 ways

By the logic given in this book, "Each arrangement is determined by where a wife is hence there are (3-1)!" ways .

Again, where have I gone wrong, please help

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    $\begingroup$ If arrangements are considered the same precisely if one can be obtained from the other by rotation, the answer is $(4)(2)$. I don't know whether you think this is correct or not. The answer is definitely not $(3-1)!$. $\endgroup$ – André Nicolas Jun 14 '15 at 13:34
  • $\begingroup$ Thanks André Nicolas . could you please also tell me why the other approach fails? $\endgroup$ – Kiran Jun 14 '15 at 13:53
  • $\begingroup$ But even with my approach, I can change the order of the first couple taken as reference which gives me 8*2 = 16, isn't it? $\endgroup$ – Kiran Jun 14 '15 at 13:58
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    $\begingroup$ To get rid of worries about rotation, let us seat $A$ at a specific chair. Then the position of $a$ is determined, There are $4$ choices for $B$, then $b$ is determined, and there are $2$ choices for $C$. $\endgroup$ – André Nicolas Jun 14 '15 at 14:02
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    $\begingroup$ Not $16$. If we take $Aa$ as a reference, we can't interchange $A$ and $a$ since such an interchange can be obtained by a rotation. $\endgroup$ – André Nicolas Jun 14 '15 at 14:21

Let us solve the general problem, since $6$ people is a very small number, and there are too many correct ways of counting. (There are also incorrect ways!)

We have $n$ couples, $\{A_1,a_1\}$, $\{A_2,a_2\}$, and so on up to $A_n,a_n$. Here we define $A_i$ to be the fatter member of couple $i$.

To make sure that we do not inadvertently double-count two arrangements that are the same under a rotation, let us seat $A_1$ at a specific chair. Then the position of $a_1$ is determined. Draw a circle with $2n$ chairs around it (a regular polygon with $2n$ vertices will also do nicely).

Now let us look at the $n-1$ chairs that go counterclockwise from the chair occupied by $A_1$ to the one occupied by $a_1$. There are $(n-1)!$ ways to decide which of the $n-1$ couples will have a member occupying these chairs. For every such choice, there are $2$ ways to decide which member of the couple will occupy the chair. That gives a total of $(n-1)!\cdot 2^{n-1}$ arrangements.

Another way: Once we have seated $A_1$ and $a_1$, there are $2n-2$ places left for $A_2$, and then the position of $a_2$ is determined. For each of these ways, there are $2n-4$ ways to choose the position of $A_3$, and then the position of $a_3$ is determined.

And so on. We get a total of $(2n-2)(2n-4)(2n-6)\cdots (2)$.

  • $\begingroup$ So the generalized answer is based on the counting (3-1)!2!2! in the example, right? $\endgroup$ – Kiran Jun 14 '15 at 14:45
  • $\begingroup$ Based on the reasoning you have explained, this will be equal to (2n-2)(2n-4)..2 = 2^(n-1) * (n-1)! so both these gives the same $\endgroup$ – Kiran Jun 14 '15 at 14:49
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    $\begingroup$ You have got it. For lots of couples, the "factorial" approach is more compact, the other approach is in my opinion more concrete. There are $2n-2$ ways to place $A_2$, and for each of these ways $2n-4$ ways to place $A_3$ all the way down to $2$ ways to place $A_n$. So we get the product you described in the comment, and it does simplify as you noted. $\endgroup$ – André Nicolas Jun 14 '15 at 15:01

If you know where the wives are seated, you know where the husbands are. There are $(3-1)!$ ways of arranging the wives around the table. You're not considering many of your eight approaches are equivalent (by rotating the people). I suggest you try and write down two "different" arrangements according to your approach, and then find they are equivalent.

  • $\begingroup$ But we are placing one pair initially and therefore all the remaining 4 positions are labeled. Therefore should be the answer? I will try marking each patter and see now. $\endgroup$ – Kiran Jun 14 '15 at 10:59
  • $\begingroup$ also see which has a similar reasoning. one of this must be wrong, but which one and why $\endgroup$ – Kiran Jun 14 '15 at 11:05

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