smc

We tackle the problem by sorting it by how many stores are involved in the transaction. 1) 2 stores are involved. There are $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ ways to choose which of the stores are involved and 2 ways to choose which store recieves the parents. $6\cdot 2=12$ total arrangements. 2) 3 stores are involved. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ ways to choose which of the stores are involved. We then break the problem down to into two subsections - when the parents and grouped together or sold separately. Separately: All children must be in one store. There are $3!$ ways to arrange this. $6$ ways in total. Together: Both parents are in one store and the 3 children are split between the other two. There are $\left(\genfrac{}{}{0ex}{}{3}{2}\right)$ ways to split the children and $3!$ ways to choose to which store each group will be sold. $3!\cdot \left(\genfrac{}{}{0ex}{}{3}{2}\right)=18$. $(6+18)\cdot 4=96$ total arrangements. 3) All 4 stores are involved. We break down the problem as previously shown. Separately: All children must be split between two stores. There are $\left(\genfrac{}{}{0ex}{}{3}{2}\right)=3$ ways to arrange this. We can then arrange which group is sold to which store in $4!$ ways. $4!\cdot 3=72$. Together: Both parents are in one store and the 3 children are each in another store. There are $4!=24$ ways to arrange this. $24+72=96$ total arrangements. Final Answer: $12+96+96=\boxed{204}$.