[Combinatorics]
Problem: Bobby, Peter, Greg, Cindy, Jan, and Marcia line up for ice cream. In an acceptable lineup, Greg is ahead of Peter, Peter is ahead of Bobby, Marcia is ahead of Jan, and Jan is ahead of Cindy. For example, the lineup with Greg in front, followed by Peter, Marcia, Jan, Cindy, and Bobby, in that order, is an acceptable lineup. Compute the number of acceptable lineups.

Solution: There are 6 people, so there are $6 !=720$ permutations. However, for each arrangement of the boys, there are $3 !=6$ permutations of the girls, of which only one yields an acceptable lineup. The same logic holds for the boys. Thus the total number of permutations must be divided by $3 ! \cdot 3 !=36$, yielding $6 ! /(3 ! \cdot 3 !)=\mathbf{2 0}$ acceptable lineups.
Once the positions of Greg, Peter, and Bobby are determined, the entire lineup is determined, because there is only one acceptable ordering of the three girls. Because the boys occupy three of the six positions, there are $\left(\begin{array}{l}6 \\ 3\end{array}\right)=\mathbf{2 0}$ acceptable lineups.

Final answer: 20