Brunaldi -- Question 2.30 Circular Permutations -- Another Participant Frees Problem Constraint 
We are to seat five boys, five girls, and one parent in a circular arrangement around a table. In how many ways can this be done if no boys is to sit next to a boy and no girl is to sit next to a girl? What if there are two parents?

I've attempted to solve the question above. I cannot solve the second part. Here is my approach to the first part:
The total number of circular permutations is obtained by multiplying together the total number of ways to arrange the boys, the girls, and the parent(s). Since no boys can be adjacent, each boy is necessarily separated by someone, either a parent or a girl.
B _ B _ B _ B _ B _
The same is true of the girls in this problem. So we place them between the boys:
B G B G B G B G B G
Assuming for now that the arrangement takes place on a straight line, since we can permute the boys and girls, we have 5! * 5! ways of arranging the boys and girls (5! for each). This over-counts the total number of arrangements, however, since given any arrangement, we can rotate the table in 5 different ways to obtain a different linear arrangement.
Thus we have $5!5!/5$ ways of arranging all the participants. The addition of one parent allows for, given any one arrangement, 10 additional ones.
Thus we have $2*5!^{2}$ ways to arrange the first setup.

What about the second? I've had no luck -- the most difficult consideration is that the introduction of a second parent frees the need of our first construction for boys and girls to interleave each other.
 A: Let us call the first parent $P$ and the second one $Q$
$\underline{Part\;1}$
Fix $P$ at the $12\; o'clock$ position as reference (for both parts)
Looking clockwise, you can either start with $B$ or $G$,
thus (arrived at a different way from your approach), ans is the same, $\;\;2\cdot(5!)^2$
$\underline{Part\; two}$
If $P$ is flanked by different genders,
all we need to do now is to fit $Q$ in any of the $11$ gaps, thus $11\cdot2\cdot(5!)^2$
But now another possibility exists, with $P$ being flanked by identical genders, say $GPG$. This will necessarily entail a $BQB$ clump in any of $4$ places.
Thus we need to further add $2\cdot4\cdot(5!)^2$ ways,
giving an overall answer of $30\cdot(5!)^2$
A: Unfortunately all the posted solutions are wrong. The correct answer is $30\cdot(5!)^2$.
Consider the following cases:
Case 1: we seat one parent, five boys and five girls as part 1 so that there are no boys sit next to a boy and no girls sit next to a girl. Then we seat the remaining parent, since there are no boys sit next to a boy and no girls sit next to a girl, the second parent can be any place among the 11 gaps. So in this case there are total $2\cdot 11\cdot (5!)^2$ possible ways.
Case 2: we again seat one parent, five boys and five girls but require that either there is exactly one pair of boys or girls sit next to each other. After that we seat the remaining parent in between this pair. For example, assume we let a pair of boys sit next to each other, then we can have following arrangement
$$
P\ G\ \bar B\ G\ B\ G\ B\ G\ B\ G,
$$
$\bar B$ denotes the pair of boy sit next to each other. Because $\bar B$ can occur in any 4 slots occupied by boys, and the arrangement of boys is irrelevant to that of girls, further, the situation when we consider girls instead of boys is symmetric, we have $2\cdot 4\cdot (5!)^2$ possible ways.
By combining two disjoint cases we have a total $30\cdot(5!)^2$ possible ways.
