Probability of men and women sitting at a table alternately

Question

I know there are already questions like this, but what I need help understanding is my book's solution in this context.

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? Write an expression that shows the probability of that happening.

I did:

• Total combinations = $16!$
• Total combinations in which women are not sitting next to each other: $$8*8*7*7*6*6*5*5*4*4*3*3*2*2 = 8!8!$$

So the answer is: $$\frac{8!8!}{16!}$$

However, my book says the solution is $$\frac{2*8!*8!}{16!}$$

Why is this? What did I do wrong?

Answer 1

The extra factor of two comes from which seats the women actually sit on. For example, if you number the seats clockwise from 1 to 16, your solution gives all cases for which the women sit on odd seats and the men sit on even seats. But the other situation is where women sit on even seats and men sit on odd seats, so there are actually twice as many solutions.

Answer 2

Your first $8$ is either the number of ways to choose the first man, or the number of ways to choose the first woman. You need to start by multiplying it by $2$, because that's the number of ways of choosing the first gender.

Alternately, consider that the first person can be anyone, ie any of the $16$ people. Then the second person must be the opposite gender, of which there are $8$ remaining, and then one of the opposite opposite gender, of which there are $7$, and then another $7$, a $6$ and so on. That's $16 \cdot 8! \cdot 7!$, which is the same thing they have (divided by 16! of course).