Four boys and 4 girls are lined up in random order. What is the probability that the boys and girls alternate Homework question and while there are very similar questions already answered my real concern is what the denominator should be in this question $8!$ or $4!$ since it is asking for the probability and not the number of possible arrangements. 
The way I read it there are 8! possible ways to randomly order the boys and girls and then in the numerator there should only be two ways to alternate boys and girls??
so my answer is $2/8!$ but I'm unsure if this is correct??
 A: There is a slight error in the reasoning, there are in fact $2\times 4!\times 4!$ ways in which they are alternated. Since there are first two choices as to which spots are occupied by the boys and which by the girls. After this there are $4!$ ways to arrange the boys and $4!$ ways to arrange the girls within their positions.
Hence we want $\frac{2\times 4!^2}{8!}=\frac{1152}{40320}$
A: Assuming $4$ identical boys and $4$ identical girls on a row, the number of permutations (i.e. the denominator of your ratio) is $\binom{8}{4}$. Among these ones, only two of them satisfy the requirement to be alternate boys/girls. It follows that the answer is
$$\frac{2\cdot 4!\cdot 4!}{8!}=\frac{24}{35}.$$
A: Alternatively, there are $\binom{8}{4}$ ways to chose the positions of the girls, and only $2$ ways of picking the locations so that they alternate with the boys, for a probability of $$\frac{2}{\binom{8}4}$$
A: There are $8!$ possible orderings of the eight people. If boys and girls alternate there are two cases.  
Case 1: First person is a boy.
In this case, there are 4 choices for the first person, 4 choices for the second person since it must be a girl, 3 choices for the 3 person, since there are 3 boys left, and so on.  Continuing in this manner we see that the total number of possibilities are $(4!)^2$.
Case 2: First person is a girl. This case is similar and so there are $(4!)^2$ orders for case 2.
Therefore the probability is $2(4!)^2/8$!.
