Four pairs of siblings, in each pair one is a boy and the other is a girl. We divide them into three groups so that each group has at least two people, but siblings do not end up in the same group. In how many ways can this be done?
According to the conditions of the task, the children can be divided into groups of either $4,2,2$ or $3,3,2$.

In the first case, the members of the 4-person group can be chosen in $2^{4}$ ways - from each family, one of the siblings can be chosen -, and the remaining four children, who are no longer siblings, can form two pairs in three ways. Therefore, in the first case, there are $3 \cdot 2^{4}=48$ possibilities.

In the second case, the first 3-person group can be formed in $4 \cdot 2^{3}$ ways, since from the four sibling pairs, we can choose three in four ways, and from these three selected pairs, a 3-person group can be formed in $2^{3}$ ways. Five remain, among whom two are siblings, and one and only one of them must be in the 2-person group, which can be formed in $2 \cdot 3$ ways. Finally, the remaining ones - the members of the third group are definitely not siblings. Therefore, in the second case, there are $\frac{4 \cdot 2^{3} \cdot 2 \cdot 3}{2}=96$ possibilities, where we divide by 2 because the order of the 3-person groups does not matter.

In total, there are $48+96=144$ possible groupings.