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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.

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    $\begingroup$ Your argument is fine. For no good reason, I would instead decide that one of the chairs is a throne, and one of the people the Queen. She sits, the remaining $n-1$ women can be permuted in $(n-1)!$ ways, and for each way the men can be permuted in $n!$ ways, for a total of $(n-1)!n!$$. $\endgroup$
    – André Nicolas
    Feb 10, 2015 at 4:07
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    $\begingroup$ What a curious way of seeing it. I'd rather think of it by fixing a place for the women, then a place for the men, and after counting all of the permutations in those places, one has to divide because rotations are or give equal permutations. $\endgroup$
    – Diego Robayo
    Feb 10, 2015 at 4:15
  • $\begingroup$ @AndréNicolas Wow that's a really easy way to think about this problem. Thanks. $\endgroup$
    – user144809
    Feb 10, 2015 at 4:21
  • $\begingroup$ It is just an alternative way. Yours was good, well explained. $\endgroup$
    – André Nicolas
    Feb 10, 2015 at 4:25

1 Answer 1

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Yes, your argument is fine. But it can also be thought of as this way. Take $n$ male, and the number of arranging them around a round table is $$(n-1)!$$ Now you are left with $n$ female but they can be placed in a total of $n!*1$ ways(arranging them by factorial and placing them in only 1 way(between the males)). Therefore the answer should be $$n!(n-1)!$$

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