# 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?

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 book's answer makes sense if the seats are labeled. – N. F. Taussig Jan 31 '16 at 16:17
• The book obviously fails to define what a "round table" should be. For another interpretation, as a cyclic graph, ignoring any "left and right", $\frac 1 2\times 4!\times 5!$ would be another answer. – Gyro Gearloose Jan 31 '16 at 17:43
• The usual convention is that seatings which can be obtained from each other by a rotation are to be considered the same. Under that convention, the number is $4!5!$. – André Nicolas Jan 31 '16 at 19:03

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.