Partition into subgroups In a class of 7 boys and 8 girls, how many ways are there to partition the class into 4 mutually disjoint unlabelled subgroups such that each of the subgroup has at least one girl and that the girls A and B will not belong to the same group?
 A: First lets see how to partition the girls. They need to be in four sets since each set must have at least one girl.
We seperate into three cases: 
Case 1 is when girl $A$ is not the only girl in her group. In this case we first seperate all girls except $A$ in $7 \brace 4$  ways(stirling coefficient of the second kind). This is because $A$ is not the only girl in her group, so every group must contain at least one of the $7$ girls that are not $A$. After doing this there will be four groups into which $A$ can be put. However one of those groups has $B$, so we can't pick that one. But there are three options. Therefore there are $3  {{7} \brace 4}$ ways to do it.
Case 2: girl $A$ is alone. Then we seperate the remaining girls in $7 \brace 3$ ways.
Therefore there are $3  {{7} \brace 4}+ {7\brace 3}$ ways to seperate the girls into $4$ subgroups.
After the girls have been seperate we proceed to seperate the boys, but this is easy because now the subgroups are "labelled" since the girls that are allready in them distinguish them. So there are $4^7$ ways to assign the boys.
Thus the final answer is $4^7(3  {{7} \brace 4}+ {7\brace 3})=(16384)(3(350)+301)=22,134,784$
