I have 2 different problems that seem identical:

8 men and 8 women are going to sit at a round table where there are 16 seats. They take their seats randomly. How many ways can the 16 seats be taken so that no 2 women are sitting next to each other?

I solved this by doing $$2 * 8!8!$$

The method I used was that $8*8*7*7*6*6*5*5*4*4*3*3*2*2*1*1= 8!8!$ is all the ways you can sit these 8 women alternating them with the 8 men.

I then multiplied by two because you have 2 possibilities: the women can be sitting at the odd numbered seats, or at the even numbered ones.

Then I have this other problem:

In how many ways can a party of 4 men and 4 women be seated at a circular table so that no two women are adjacent?

(I took the problem from here: http://www.imsc.res.in/~kamalakshya/cupboard/comb_mag.pdf)

And the solution, according to the website, is:

Answer: The 4 men can be seated at the circular table such that there is a vacant seat between every pair of men in (4-1)! =3! Ways. Now 4 vacant seats can be occupied by 4 women in 4! Ways. Hence the required number of seating arrangements = 3!4! = 144

However, I can't seem to use the method from the 1st problem on this one. All the possible combinations of men and women, sitting alternated with men, would be $4!4!$ Then, multiplying by 2 the solution would be $2 * 4!4! = 1152$

If I use the method from the answer of the 2nd problem with the 1st one, I get $7!8!$, which is different from $2 * 8!8!$

My question is, how can this be? Is there any difference between these 2 problems that I am not seeing?

EDIT: I believe the 1st solution is correct. Check here Probability of men and women sitting at a table alternately

  • $\begingroup$ You are assuming that the seats are labeled. The answer to the second problem assumes the seats are not labeled and takes rotational invariance into account. $\endgroup$ – N. F. Taussig Oct 16 '16 at 1:29
  • $\begingroup$ Take into account circular permutations. The 1st solution is wring 2nd is correct. $\endgroup$ – Navin Oct 16 '16 at 1:30
  • $\begingroup$ @navinstudent math.stackexchange.com/questions/1969572/… $\endgroup$ – SilenceOnTheWire Oct 16 '16 at 1:32
  • $\begingroup$ @N.F.Taussig That makes sense, but how do you infer which method to use from reading the problems, without looking at their solutions? Neither problem mentions anything about the seats being labeled. $\endgroup$ – SilenceOnTheWire Oct 16 '16 at 1:52
  • $\begingroup$ If the seats are not described as labeled, I would assume that they are unlabeled and that rotational invariance applies, as in the second problem. $\endgroup$ – N. F. Taussig Oct 16 '16 at 2:57

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