In how many ways can 5 men and 5 women sit at a round table such that no 2 persons of the same gender sit next to each other? The book's answer is $2\times 5! \times 5!$ Why is it not $2\times 4! \times 5!$ ? Is the circular arrangement the same as a line one in this case?
The circularity of the table is just to influence what seats are next to one another (note that if they sit on a line, then it doesn't have to be every other man and woman, since the end seats aren't neighbours). There is still one seat that is "seat number one", "closest to the door", or "on the north side of the table", or whichever other distinguishing feature you would prefer.
Let's mark one seat as "Seat $1$" and then number the seats, incrementing by $1$ as we go counter clockwise and stopping when we get to Seat $1$ again. Seat $1$ can either be a man or a woman, which is $2$ possibilities. Then, all of the odd seats must have the same gender as Seat $1$ and all of the even seats must have the opposite gender as Seat $1$. However, we can rearrange the people in the odd seats and since there are $5$ odd seats ($1, 3, 5, 7, 9$), this has $5!$ possibilities. Similarly, we can rearrange the people in the even seats and since there are $5$ even seats ($2, 4, 6, 8, 10$), this also has $5!$ possibilities.
Therefore, we multiply all of the possibilities together to find that there are $2*5!*5!$ ways to arrange all of the people so that no two adjacent people have the same gender.