AMC10 2018 B

Answer (D): Let $X$, $Y$, and $Z$ denote the three different families in some order. Then the only possible arrangements are to have the second row be members of $XYZ$ and the third row be members of $ZXY$, or to have the second row be members of $XYZ$ and the third row be members of $YZX$. Note that these are not the same, because in the first case one sibling pair occupy the right-most seat in the second row and the left-most seat in the third row, whereas in the second case this does not happen. (Having members of $XYX$ in the second row does not work because then the third row must be members of $ZYZ$ to avoid consecutive members of $Z$; but in this case one of the $Y$ siblings would be seated directly in front of the other $Y$ sibling.) In each of these 2 cases there are $3! = 6$ ways to assign the families to the letters and $2^3 = 8$ ways to position the boy and girl within the seats assigned to the families. Therefore the total number of seating arrangements is $2 \cdot 6 \cdot 8 = 96$.