Two circular seating arrangements are considered to be distinguishable if the relative order of the people differs.
In how many ways can six men sit at a round table?
Method 1: Consider the diagram below.
We can describe a particular seating arrangement by describing the order of the men as we proceed counterclockwise (anticlockwise) around the circle. Notice, however, that we can describe this particular seating arrangement in six ways, depending on whether we start with A, B, C, D, E, or F. Thus, this particular circular arrangement corresponds the six linear arrangements ABCDEF, BCDEFA, CDEFAB, DEFABC, EFABCD, FEABCD. More generally, each circular arrangement of the six men corresponds to six linear arrangements, corresponding to the six possible starting points as we proceed counterclockwise around the circle. Since six men can be arranged in a row in $6!$ ways, there are
$$\frac{6!}{6} = 5!$$
distinguishable circular arrangements of the men.
More generally, we can arrange $n \geq 1$ distinct objects in a circle in
$$\frac{n!}{n} = (n - 1)!$$
ways since each circular arrangement corresponds to $n$ linear arrangements, one for which each of the $n$ starting points as we proceed counterclockwise around the circle.
Method 2: We consider arrangements relative to the first man we sit at the table.
We seat man A. We use him as our reference point. As we proceed counterclockwise around the circle, we can seat the remaining five men in $5!$ orders relative to man A.
More generally, given a set of $n$ distinct objects, we can arrange them in $(n - 1)!$ orders as we proceed counterclockwise around the circle relative to the first object we place on the circle.
In how many ways can six men and five women dine at a round table if no two women sit together.
We can seat the men in $5!$ distinguishable ways. This create six spaces, one to the immediate right of each man. To ensure that no two of the women sit in adjacent seats, we must choose five of those six spaces in which to insert a woman. Once those spaces have been selected, we can arrange the five women in those spaces in $5!$ orders. Hence, the number of possible seating arrangements is
$$5! \binom{6}{5} \cdot 5! = 5!P(6, 5)$$
Given it is a circular table, when you are seating women first or men first, you need to fix one of the positions and make all arrangements relative to that (assuming rotational symmetry).
If so, there are only $4!$ ways to seat $5$ women leaving one seat in between each of them for men. Then there are $5!$ ways to seat men in those $5$ empty seats. Now you have $10$ places where you can seat the dog.
So total number of arrangements are $4! \times 5! \times 10 = 2 \times (5!)^2$.
Another way to look at this is to let the dog take the first seat as it wishes. Now we go clockwise seating women and men in alternate seats. There are $(5!)^2$ ways to do that starting with a woman as women can be arranged in $5!$ ways and so can be men. But we can also start with a man. So we have another $(5!)^2$ arrangements - in total $2 \times (5!)^2$ arrangements.
Best Answer
Let Amanda be one of the women. We seat Amanda first. This determines which $k - 1$ seats are available to the other women. The remaining $k - 1$ women can be seated in these seats in $(k - 1)!$ ways as we proceed clockwise around the circle relative to Amanda. The men can be seated in $k!$ ways as we proceed clockwise around the circle relative to Amanda. Hence, the number of permissible seating arrangements is $$(k - 1)!k!$$
Observe that if there is only one man and one woman at the table, this answer yields $1/2$ seating arrangements.
What went wrong?
Suppose Amanda is one of the women, and she is seated first. Since the men and women must alternate, seating Amanda determines in which $k$ seats the women will sit. Hence, there are only $k$ rotations that preserve a given arrangement of the women. Moreover, once the women are seated, the men cannot be rotated without changing the relative order of the people in the seating arrangement. Hence, the number of distinguishable seating arrangements is $$\frac{k!k!}{k} = (k - 1)!k!$$