Doubt with probability of men and women sitting at a table alternately I just have a question regarding this existing question. > Probability of men and women sitting at a table alternately As specified there, > There is a round table with 16 seats. 8 men and 8 women are going to sit at this table. What is the probability of the 16 seats being occupied so that none of the women sit next to another woman? The answer provided is $\frac{2.8!.8!}{16!}$ However I have a doubt in here, that I'm asking. How the total combinations is becoming 16! ? As the famous (n-1)! rule, shouldn't be it 15! ? And for man and woman sitting alternatively, why is that I'm not fixing the men in (8-1)! = 7! and then arranging the women in remaining 9 places in $^9P_8$ ways? By that the probability becomes $\frac{^9P_8.7!}{15!}$ . Please clarify me why my reasoning is wrong? Thanks in advance.

Re your query about the famous $(n-1)!$ formula, I have a few points to make:

* Here we want the **probability** , so it doesn't really matter whether we take the seats to be unnumbered (which we normally do unless it is explicitly stated otherwise) or take them as numbered.

* Taking them as numbered makes the computations symmetrical here, but we must (and do) get the same answer if we treat them as unnumbered.

* Seat the women in $7!$ ways, and the men in the $8$ spaces in between in 8! ways, to get $Pr = \frac {7!8!}{15!}$