\centerline{\bf Linear orders, cyclic orders and
permutations\strut}
\centerline{Miguel Carri\'on \'Alvarez\strut}
\proclaim 1, 2. Generating functions.
The empty set is vacuously ordered in one way; a $1$-element set is
trivially ordered in~$1$ way; and an $n$-element set can be linearly
ordered by choosing a maximal element in one of~$n$ ways and linearly
ordering the remaining $(n-1)$-element subset. Hence, $|L_n|=n!$ and
$$
|L|(z)=\sum_{n\ge 0}{|L_n|\over n!}z^n=\sum_{n\ge 0}z^n={1\over 1-z}.
$$
The empty set admits no cyclic orders. If~$\{x_1,x_2,\ldots,x_n\}$ is
an~$n$-element set, the
map
$$
x_1