Probability of alternating alternating positions I was given the following questions:

I am stuck on question b.
The probability is defined as $P(E) = \frac{n(E)}{n(S)}$
I believe $n(s) = 11!$ (or is it $\frac{11!}{(6!5!)}$)
I am stuck on solving $n(E)$
The book's answer is $P(E) = 0,02$
 A: Lining up $6$ (indistinguishable) men and $5$ (indistinguishable)
women can be done in $\binom{11}{5}=\binom{11}{6}$ ways.
Lining up $6$ (indistinguishable) men and $5$ (indistinguishable)
women under the extra condition that men and women are standing in
alternate position can be done in $1$ way.

Lining up $6$ (distinguishable) men and $5$ (distinguishable) women
can be done in $11!$ ways.
Lining up $6$ (distinguishable) men and $5$ (distinguishable) women
under the extra condition that men and women are standing in alternate
position can be done in $5!6!$ way.

The methods yield the same probability: $$\binom{11}{5}^{-1}=\frac{5!6!}{11!}$$
A: You can choose $n(S)$ to be either $11!$ or $\frac{11!}{6!5!}$, but then you have to compute $n(E)$ in the same way.

If we let $n(S)$ be $11!$, then that means we consider the men and women distinguishable from each other. Since "alternating" forces the arrangement to be "man, woman, man, woman, ..., woman, man," the only remaining thing to count is how the men are lined up relative to each other and how the women are lined up relative to each other.

 $n(E) = 5!6!$.


If we instead let $n(S)$ be $\frac{11!}{5!6!}$, then that means we consider all men to be indistinguishable and all women to be indistinguishable. In that case, there is only one way to arrange them in the alternating fashion. Note that the probability is still the same as the one obtained in the approach above.
A: Obviously, since it's alternating line, the men are on the odd spots and the women are on the even spots, so just permutate over men and set the up with that order on the odd spots, and same for the women with the even spots.
which gives us the answer $6!5!$ and the probability $$\frac{6!5!}{11!}$$
A: The only way for the men and women to be standing in alternate positions is
$$
M_1W_1M_2W_2M_3W_3M_4W_4M_5W_5M_6
$$
The 6 men can be arranged in $6!$ ways and the women in $5!$. So there are a total of $6! 5!$ permutations. Note that we multiply because they are independent of one another.
So
$$
P(E)=\frac{n(E)}{n(S)}=\frac{5!6!}{11!}
$$
