There are two main cases to consider:
Case 1: Each row has exactly one child from each family.
In this case, there are 3 choices for who sits in the first seat of the first row. Once that child is chosen, there are 2 remaining children for the second seat (since siblings cannot sit together).
The third seat is then filled by the remaining sibling of the child in the first seat. Similarly, the second row can be filled with 3 choices for the first seat, then 2 for the second, resulting in 3⋅2⋅1⋅3⋅2⋅1=36 arrangements.
Case 2: One row has two children from the same family.
Here, we can further divide into two subcases:
Subcase 2a: The first child in the first row is a sibling of the first child in the second row.
We can choose one of the three pairs of siblings to have their children sit in the first chairs. There are then 2 ways to order those siblings within the pair. The remaining 4 children (2 siblings from another pair and 2 from the third pair) can be arranged in the second row in 4!=24 ways.
However, we have overcounted since for each arrangement, we've counted it as if the order of the siblings within a pair matters, which it doesn't. Therefore, we must divide by 2!⋅2! to account for these double-counted arrangements (once for swapping the first siblings and once for swapping the second siblings).
This gives us $ \dfrac{3 \cdot 2 \cdot 24}{2! \cdot 2!} = 36$ arrangements.
Subcase 2b: The first child in the first row is NOT a sibling of the first child in the second row.
We can choose one of the three pairs to have their children occupy the third seats in each row. The remaining 4 children can then be arranged in the first row in 4! ways. Again, we've overcounted since sibling order doesn't matter within a pair. Dividing by 2!⋅2! gives us $ \dfrac{3 \cdot 24}{2! \cdot 2!} = 36$ arrangements.
Since Cases 1 and 2 are mutually exclusive, to find the total number of arrangements, we simply add the number of arrangements from each case: 36+36+36=108
Therefore, the answer is 108.
!
