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.

  • $\begingroup$ To get insight in what's wrong in your reasoning it might help to do the same thing with smaller numbers like $2$ men and $2$ women. Things like picturing and counting are made more easy then. $\endgroup$
    – drhab
    Commented Oct 19, 2016 at 10:29
  • $\begingroup$ if you seated men then women, the first man can sit in 16 seats due to symmetry, but then to create the man/woman pattern he has started, there are then only 7 seats for the next man, or if it was one of the 8 women, 8 seats - in your quoted answer, this is represented as '2' since the first man chooses one of two parities - for random ordering, there has to be 16! arrangements - my reasoning is that it is $16 7! 8! / 16! $ which equals their answer $\endgroup$
    – Cato
    Commented Oct 19, 2016 at 11:56

3 Answers 3


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 $Pr = \frac {7!8!}{15!}$

  • $\begingroup$ Nothing like this! What a great explanation! :) Just made my day. Lingered so much with this question. Thanks a ton! $\endgroup$
    – lu5er
    Commented Oct 19, 2016 at 11:37
  • $\begingroup$ That I was confusing was 9 places between 8 women, Just $^8P_8$ as you said. Thanks again! Loving this subject. $\endgroup$
    – lu5er
    Commented Oct 19, 2016 at 11:41
  • $\begingroup$ You're welcome ! Glad you are loving this subject ! $\endgroup$ Commented Oct 19, 2016 at 11:55
  • $\begingroup$ Nice that you helped the OP out of lingering! $\endgroup$
    – drhab
    Commented Oct 20, 2016 at 8:45

You can start by giving each chair a number: $1,2,\dots,16$, keeping in mind that two chairs are next to each other if their numbers are consecutive of are $1$ and $16$.

That gives $16!$ possibilities in total for sitting down without further conditions. Under the extra condition that the women (ladies first!) do not sit next to each other they are forced to possess the chairs with even numbers ($8!$ possibilities) or the chairs with odd numbers ($8!$ possibilities). Then $8$ chairs are left for men ($8!$ possibilities).

  • $\begingroup$ But it is a circular permutation, not linear one, right? Then how can it be 16! ? $\endgroup$
    – lu5er
    Commented Oct 19, 2016 at 10:28
  • $\begingroup$ There is nothing wrong here with making a row of $16$ chairs, then to sit down, and then to place the chairs back around the table. $\endgroup$
    – drhab
    Commented Oct 19, 2016 at 10:32
  • $\begingroup$ Yes, that's correct! but in case of circular permutation, bgbg is same as gbgb, isn't is? Because the relative order does not matter. $\endgroup$
    – lu5er
    Commented Oct 19, 2016 at 10:36
  • $\begingroup$ Then when to use (n-1)! in round tables and when to use n! by stretching them in a single row? $\endgroup$
    – lu5er
    Commented Oct 19, 2016 at 10:37
  • $\begingroup$ It is - I think - not so easy to give a general answer to that. In every case I meet I must puzzle about that myself (is the "roundness" of the table relevant here, or not?). I am always trying to make it as easy as possible. $\endgroup$
    – drhab
    Commented Oct 19, 2016 at 10:42

To answer why $\frac{2*8!*8!}{16!}$ we can first answer how many different combination to place the 16 men and women with no 2 women sitting to another.

beacause its a round table, it is a convention that a rotation of the table isn't consider as a different combination. so you can name any sit that you want as sit number 1, and the rest will order 2-16. now because we have 8 women out of 16 you must place all the women in the odd sits or the even (which means 2 combinations).

After choosing odd or even sits, you can place the 8 women in any order that you want in those sits. The number of combinations for placing 8 people is 8!.

As for the men after placing the women you have 8 sits left, and again you can sit them in any order that you want, as they will be sitted in any other sit - so again you have 8! combinations.

adding all up you have 2*8!*8! different combinations.

Because the total combinations for placing 16 people in a circle is 16! you get that the probability for a random ordering of the men and women will be as asked is $\frac{2*8!*8!}{16!}$.

The total number of combinations is 16! because the convention that the rotation of the table doesn't count as a different combination (it is just different point of view). so the total combination of placing people around a table is the same as placing them in a line.

  • $\begingroup$ Nice explanation. :) Got it clearly now! $\endgroup$
    – lu5er
    Commented Oct 19, 2016 at 12:58

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