Why does the $(n-1)!$ rule not work in all cases of circular permutations? 
$5$ Boys and $5$ girls sit alternatively around a round table. In how many ways can this be done?

I solved it like this : $5$ boys can be arranged in $(5-1)!$ ways. After that the $5$ girls can be arranged in the gaps in $(5-1)!$ ways. So, the answer should be $4!×4!$ but the actual answer is $4!×5!$. After seeing the answer I can guess that they have considered $5$ instead of $(5-1)$ in any one of the cases. I have learnt that number of circular arrangements = $(n-1)!$. So, why did I get wrong answer?
EDIT

rotating each child one place to the left does not produce a seating arrangement that will be counted again, simply because now the girls are sitting where we sat the boys, and vice-versa. But if we rotate everyone 2, 4, 6, or 8 seats to their left, then we will get another seating arrangement that will be counted again.

Does that mean that if for example, $5$ men, $5$ women and $5$ children are to sit alternately then the answer should be $\frac{5!×5!×5!}{5×3/3}$ because neither can we rotate each member to the left by one place nor by two places? So does that mean that in this type of questions, we have to group the objects and then divide the result with total number of groups?
 A: The circular arrangement formula takes into account that you can rotate the positions, and nothing of substance changes. If you're putting 5 people on a circular table, you get $5!$ permutations, divided by $5$ to account for the fact that we can rotate the seating arrangement $5$ different ways without substantial change to the arrangement.
That is, when you count the placements as $5!$, you are over-counting by a factor of $5$, because each of the $5!$ placements can be rotated to $4$ other seating arrangements.
In your case, you seat $5$ girls and $5$ boys alternately. Without accounting for rotations, there are $5! \times 5!$ ways of seating the children. But then we must account for rotations. How many other equivalent seating arrangements can we form by rotating the seats?
Note that rotating each child one place to the left does not produce a seating arrangement that will be counted again, simply because now the girls are sitting where we sat the boys, and vice-versa. But if we rotate everyone $2$, $4$, $6$, or $8$ seats to their left, then we will get another seating arrangement that will be counted again. This tells us that by computing $5! \times 5!$, we have again over-counted by a factor of $5$, so the result is
$$\frac{5! \times 5!}{5} = 5! \times \frac{5!}{5} = 5! \times 4!.$$
By computing $4! \times 4!$, you are implicitly assuming that you can rotate the boys and girls independently. This isn't really allowed in the problem, as it would produce different seating arrangements.
For example, if we had the cyclic order
$$\text{matt}, \text{hannah}, \text{charlie}, \text{elizabeth}, \text{warren}, \text{jenny}, \text{peter}, \text{veronica}, \text{cameron}, \text{celia},$$
then we get an equivalent order by shuffling everyone $2$ spots to the left:
$$\text{charlie}, \text{elizabeth}, \text{warren}, \text{jenny}, \text{peter}, \text{veronica}, \text{cameron}, \text{celia}, \text{matt}, \text{hannah}.$$
Everyone still has the same person to their left, and to their right. On a circular table, nobody would know the difference. But, if we were to apply the $5 \times 5$ independent rotations of the girls and the boys, we could rotate just the girls one spot to the left:
$$\text{matt}, \text{elizabeth}, \text{charlie}, \text{jenny}, \text{warren}, \text{veronica}, \text{peter}, \text{celia}, \text{cameron}, \text{hannah},$$
and we get a totally different order. Elizabeth used to be next to Charlie and Warren, but now she's next to Matt and Charlie (and Charlie is sitting on the opposite side of her to where he previously sat). As far as the table's participants are concerned, this is a different configuration.
