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.
Best Answer
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 $30\cdot(5!)^2$