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I always get confused with combination questions that involve round tables? If there were $4$ women and $3$ men sitting at a round table with no restrictions, how many possible combinations would there be if we just looked at the people's genders

For example 'WWMWMMW' would count as one seating even though the $4$ women can rearrange themselves in many ways in those $4$ women spots.

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  • $\begingroup$ Do you distinguish between constellations which differ by a cyclic permutation? $\endgroup$ – Peter Oct 5 '15 at 10:10
  • $\begingroup$ Welcome to Math.SE! Did you see that there are already quite a number of questions on the "placing people around a table" problem? It would be very helpful for you to read them and to explain in your question what it is that you still do not understand. $\endgroup$ – Hrodelbert Oct 5 '15 at 10:20
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If the chairs would have a number then the answer would be $\binom73=35=\binom74$. You just pick out $3$ numbered chairs for the men (or $4$ for the women if you like).

If the chairs are not numbered then each distinct arrangement will be counted $7$ times by the method above. This because e.g. choice $123$ is in that case actually the same as $234$,$345$,...,$671$,$712$.

So in that case there are $\frac17\binom73=5$ essentially distinct arrangements.

Be cautious, though. Luckily $7$ is a prime wich ensures that the overcounting is determined by factor $7$. This will not be so if $7$ is not a prime. If e.g. $2$ men and $2$ women must be have a seat then there are $2$ possibilities (the men sit next to each other or not), but $2\neq3=\frac12\binom42$.

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There are two groups so, 2 groups can be arranged in 1 way on table(Note the idea here, try to figure out that if they were sitting on line then ans would have 2! = 2 but for sitting on round line we subtract 1 & then calculate factorial)( Since on both the cases whether women group sits facing west or men group facing west, they both come to the left & right of each other), but if seated in group.
Now 3 men can be arranged in 3P3 ways i.e. 3! = 3X2 = 6 ways with each other
also Now 4 women can be arranged in 4P4 ways i.e. 4! = 4X3X2 = 24 ways with each other

if they are mixed with each other on line then 7P7 ways = 7! = 5040 ways


Now if they are sitting on round table then just subtract 1 & calculate factorial, So (7-1)! = 720 ways(ans)

If you are trying to figure out how formula works then this will waste lots of time. So don't get confused & just subtract 1 before calculating normal permutation. That's it.

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