In how many ways can $3$ couples sit around a round table such that men and women alternate and none of the couples sit together?
My solution goes like this:
We denote the $3$ couples by the symbols $(B_1,G_1),(B_2,G_2)$ and $(B_3,G_3)$ where each such ordered pair i.e $(B_i,G_i)$ ( where $B_i's$ and $G_i's$ are men and ladies respectively) constitute a couple. Now, we consider the positions of $B_1$ that are possible in the given arrangement. We observe that $G_1$ and $G_2$ are the only candidates that can sit on either side of $B_1.$ We have two cases:
Case 1: This is the case when the arrangement of $B_1,G_2,G_3$ are as shown below in the figure:
$$G_2,B_1,G_3$$
Now, neither $B_2,B_1$ can sit on the other side of $G_2$ as $B_1$ is already seated and $B_2$ cant sit beside $G_2$ as otherwise the given condition in the question would be violated. So, only $B_3$ can be seated on the other side of $G_2.$ The arrangement would then proceed to look like this:
$$B_3,G_2,B_1,G_3$$
We continue our reasoning. Now, similarly, only $B_2$ can be seated on the other side of $G_3.$ The arrangement would now look like:
$$B_3,G_2,B_1,G_3,B_2$$
We must not forget, that the couples are seated on a round table and hence there is only one possible position for placing $G_1$, i.e between $B_3$ and $B_2.$ Placing $G_1$ there, we find, indeed it is a valid arrangement. Our final arrangement looks like this:
$$B_3,G_2,B_1,G_3,B_2,G_1$$
We now proceed to develop Case 2.
Case 2: In this case, we assume $B_1,G_2,G_3$ are arranged like this, as shown below in the figure:
$$G_3,B_1,G_2$$
Reasoning similarly, in the previous analogous case, we find, $B_2$ and $B_3$ must be seated in the vacant positions beside $G_3$ and $G_2$ respectively. The arrangement would begin to look like this:
$$B_2,G_3,B_1,G_2,B_3$$
Again, we note that if $G_1$ is placed in between $B_2$ and $B_3$ then it becomes a valid arrangement. So, the final arrangement looks like this:
$$B_2,G_3,B_1,G_2,B_3,G_1$$
But then again, as the arrangements are done on a round table, we note that both the final arrangements of Case 1 and Case 2 i.e $$B_3,G_2,B_1,G_3,B_2,G_1$$ and $$B_2,G_3,B_1,G_2,B_3,G_1$$ are the same or rather equivalent. So, wherever we place $B_1$ to start the process, it doesn't matter as we will get only one possible arrangement only. So, the number of possible arrangement is $1.$
Now, I am totally confused whether my solution is correct or not? This is because the answer given is $2$ instead of $1$ possible arrangement(s). I seem not to get a hold of this dilemma.
EDIT: As it seems from the comments, in this problem, reflected arrangements are considered to be different in a round table arrangement. While, in my solution, I assumed them to be equivalent. This was the cause of disparity between the answers. But, now I want to know how to understand whether two arrangements are reflections of each other or not. As because, in this problem I had no idea that the two possible arrangements I considered in the two cases are mere reflections of each other. To be precise: I am confused and I want to know how to find all possible reflected arrangement for a given arrangement. I seem to have no idea about these terminologies /results, or better to say: the notion of "reflections of a particular arrangement" was completely unknown/new to me up until this day.
If you want to solve these type of questions without writing the cases as you did, i suggest you using principle of inclusion-exclusion
Our logic is that ( the number of all alternating arrangements) - (the number of alternating arrangements that contain at least one adjacent couple).
Now, i assume that you know how to apply P.I.E,so
$$2!\times 3! - \bigg(\binom{3}{1}\times2\times 4-\binom{3}{2}\times 2\times3+ \binom{3}{3}\times 2\times 2 \bigg)=12-24+18-4=2$$
$2!\times 3! \rightarrow$ Place firstly men by $(3-1)!=2!$ ,and after that, place the women the gaps between men by $3!$ ways.
$\binom{3}{1}\times2\times 4 \rightarrow$ Firslty select one couple to make them together. After that , place $2$ men and $2$ women around circle by $(2-1)!\times 2!=2$ ways.Now, we obtained $4$ gaps to place this couple and their inter arrangements can be done only one way, because when this couple place an suitable place, they should obey alternating sitting rule.
$\binom{3}{2}\times2 \times3 \rightarrow$ Select two couple by $C(3,2)$ to place them together. When we place a man and a woman, we have $2$ gaps to place first couple.After that ,now we have $3$ gaps to place the other couple.Again, the inter arrangements of couples does not important ,because there will be only one sitting rule to obey alternating sitting rule.
$\binom{3}{3}\times 2\times 2 \rightarrow$ All couples are together $C(3,3)$. Then, if we think each couple as a single object, these $3$ couples can be placed $(3-1)!=2!$ ways.However, if we place the first couple as $(W,M)$ according to clockwise, then the second couple will also be placed as $(W,M)$ and so on, but if we place the first couple as $(M,W)$ according to clockwise, then the second couple will also be placed as $(M,W)$ and so on.Hence, the inter position of couples in first arrangement is important and there are $2$ distinct inter sitting for couples.