8. Six people, all of different weights, are trying to build a human pyramid: that is, they get into the formation
$$
\begin{array}{c}
\text { A } \\
\text { B C } \\
\text { D E F }
\end{array}
$$

We say that someone not in the bottom row is "supported by" each of the two closest people beneath her or him. How many different pyramids are possible, if nobody can be supported by anybody of lower weight?
Answer: 16
Solution: Without loss of generality, let the weights of the people be $1,2,3,4,5$, and 6 . Clearly we must have $A=1$. Then, equally clearly, either $B$ or $C$ must be 2 . Suppose $B=2$ : Then either $C$ or $D$ must be 3 . If $C=3$, we have 3 ! $=6$ possibilities to fill the bottom row. If $D=3$, then $C=4$ and we have $2!=2$ possibilities to fill $E$ and $F$. Altogether there are $6+2=8$ possibilities in this case.
Suppose $C=2$ : then, similarly, there are 8 possibilities here.
Altogether there are $8+8=16$ possibilities.