jmc

In all, disregarding the gender rule, there are $$\left(\genfrac{}{}{0ex}{}{12}{4}\right)\left(\genfrac{}{}{0ex}{}{8}{4}\right)=\frac{12\cdot 11\cdot 10\cdot 9\cdot 8\cdot 7\cdot 6\cdot 5}{4\cdot 3\cdot 2\cdot 4\cdot 3\cdot 2}=34650$$ways of assigning the teams. We will count the number of ways a team can have all boys or all girls and subtract from this total.

There are 2 choices for the violating gender and 3 choices for the violating color. Once these are picked, there are $\left(\genfrac{}{}{0ex}{}{6}{4}\right)=15$ ways to choose the violating team, and $\left(\genfrac{}{}{0ex}{}{8}{4}\right)=70$ ways to pick the other two teams, for a total of $2\cdot 3\cdot 15\cdot 70=6300$ ways to choose a violating team. However, this procedure double-counts the assignments that make one team all girls and another all boys. There are 3 choices for the girls team and then 2 choices for the boys team, and ${\left(\genfrac{}{}{0ex}{}{6}{4}\right)}^2=225$ ways to choose the teams, for a total of $2\cdot 3\cdot 225=1350$ double-counted arrangements, leaving $6300-1350=4950$ ways to make a team all girls or all boys. Subtracting this from the total, we get $34650-4950=\boxed{29700}$ ways for the coach to assign the teams.