Seating 10 people in a circular table In how many ways can 10 people be seated in a circle if:
1) there are 5 men and 5 women and no two men and no two women can sit next to each other?
2)six of them are men and they must sit next to each other?
3) there are four married couples and each husband must sit next to his wife?
I gave a try at solving each of there but am no sure if I did it correctly because I am not sure how to take the circular table into account when counting:
1) first we choose the 5 men or women out of the 10 people $\binom{10}{5}$, then there is $5!$ ways of arranging them, then the remaining people have $5!$ ways of being arranged so my answer would be  $\binom{10}{5}$$(5!)^2$
2) choose 6 men out of 10 people $\binom{10}{6}$, then there is $6!$ arrangements for them to sit next to each other, then the remaining women can be seated in $4!$ arrangements so my answer would be 
$\binom{10}{6}$$6!$$4!$
3)take each couple as one person, so now there is only 6 people. There is $6!$ different arrangements for everyone to be seated then the two people that are not in a couple can be arranged in $(2!)^2$ ways so my answer would be $6!(2!)^2$
 A: 1)  Since two arrangements are considered equivalent if they only differ by a rotation, let us imagine one of the people is the Queen, and one of the chairs is a throne.  The Queen sits down on the throne. Now the places occupied by women are determined, but the women can be permuted in these positions in $4!$ ways. Then the men can be inserted in the empty spots in $5!$ ways, for a total of $4!5!$.
2) Again there is a throne, and we can suppose that there is a man on the throne, and all the men are in the five chairs counterclockwise from the throne. The men can be permuted in $6!$ ways, and the women inserted in the empty spots in $4!$ ways.
3) Again the Queen (who is married) sits on the throne. Her husband has $2$ choices. Now there are $3$ couples left, and $2$ singles. We can think of these as $5$ objects, which can be permuted in $5!$ ways, and then, as in your solution, the couples can have members switched in $2^3$ ways, for a total of $(2)(5!)(2^3)$.
Remark: We can solve all of these problems by considering the chairs to be labelled, and count the possible seatings. Then divide the number of labelled seatings by $10$ to deal with the fact that two seatings that are rotationally equivalent are considered the same. But the trick used in the solutions often gives a more concrete visualization.
