# If there are $n$ number of males and $n$ number of females, how many ways to seat them about a round table if sexes alternate?

If there are $n$ number of males and $n$ number of females, how many ways can they be seated about a round table if the sexes alternate and arrangements are considered the same when one can be obtained from the other by rotation?

My approach:

Since there are $n$ number of males and $n$ number of females, there are $2n$ number of people in total. If I seat them around a circular table, I can choose to first sit a male or a female, and sit last the sex that did not sit first (i.e. for $n=2$ I can order them as $m_{1},f_{1},m_{2},f_{2}$ or $f_{1},m_{1},f_{2},m_{2}$, where $m$ is a male, $f$ is a female).

If I count the way people could be chosen from one way (CW or CCW), by the multiplication principle there are $n$ ways to select the first person of one sex and $n$ ways to select the first person of the other sex, then $(n-1)^2$ to select the second people of both sexes, then $(n-1)^3$ to select the the third people of both sexes, all the way to the last people $(n-(n-1))^2$.

So there are $n^2*(n-1)^2*(n-2)^2*...*(n-(n-1))^2$ ways. Since there are $n$ similar linear arrangements that could be obtained from rotation, there are $$\frac{n^2*(n-1)^2*(n-2)^2*...*(n-(n-1))^2}{n}=n*(n-1)^2*(n-2)^2*...*(n-(n-1))^2$$ arrangements so that the sexes alternate and arrangements that are the same by rotation are not counted.

Is this correct? Thanks.