The "ring" question asks "how many ways are there to put 6 people in a ring", where two "ways" are identical if each individual has the same left and right neighbours.
The "table" question asks "how many ways are there to put 6 people at a table", where two "ways" are identical if each individual has the same left and right neighbours and the same place at the table.
It is easy to see that since the possible rotations are 6, the answer to the first question will be 6 times the answer to the second question, and indeed 6! = 6 x 5!.
Concerning the "necklace" problem, the difference from the "ring" problem is that you cannot put people upside down without noticing - while you can to so with a necklace, obtaining a pattern which is symmetric. So the number of combination is halved, because two symmetric combinations are considered as one.
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
$$5! \binom{6}{5} \cdot 5! = 5!P(6, 5)$$
Best Answer
First seat the $6$ men. If they were seated in a row, there would be $6!$ possible permutations of them. However, they’re actually seated around a circular table, which changes things. The six permutations $ABCDEF,BCDEFA,CDEFAB,DEFABC,EFABCD$, and $FABCDE$ are the same when the men are sitting in a circle: the only differences is where in the circle you started the list. Thus, each arrangement of the men around the table corresponds to $6$ permutations in a straight line, and there are therefore only $\frac{6!}6=5!$ possible arrangements of the men around the table. (This is a useful general fact: there are $(n-1)!$ circular permutations of $n$ things.)
Now we seat the $5$ women. There are $6$ positions available between adjacent men, so we have $6$ possible ways to seat the first woman. If we put another woman between the same two men, she’d be sitting next to the first woman; we don’t want that, so we have to seat her in one of the other $5$ gaps between adjacent men. That leaves $4$ open gaps for the third women, $3$ for the fourth woman, and $2$ for the last woman, so altogether we can seat the women in $6\cdot5\cdot4\cdot3\cdot2=6!$ different ways.
Combining the two results, we see that the $11$ people can be seated in $5!\cdot6!$ different orders around the table if no two women sit next to each other.