Married couples around a table(2) 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
 A: 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 $(2n-2)(2n-4)(2n-6)\cdots (2)$.
A: 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.
