23. Two numbers $x, y \in\{1, \ldots, 2 n\}$ will be called twins if $|x-y|=n$. Then the set $\{1, \ldots, 2 n\}$ splits into $n$ pairs of twins. A permutation $\left(x_{1}, \ldots, x_{2 n}\right)$ of this set is said to be of type $T_{k}$ if $\left|x_{i}-x_{i+1}\right|=n$ holds for exactly $k$ indices $i$ (thus a permutation of type $T_{0}$ contains no pairs of neighboring twins). Denote by $F_{k}(n)$ the number of $T_{k}$-type permutations of $\{1, \ldots, 2 n\}$. Let $\left(x_{1}, \ldots, x_{2 n}\right)$ be a permutation of type $T_{0}$. Removing $x_{2 n}$ and its twin, we obtain a permutation of $2 n-2$ elements consisting of $n-1$ pairs of twins. This new permutation is of one of the following types: (i) type $T_{0}: x_{2 n}$ can take $2 n$ values, and its twin can take any of $2 n-2$ positions; (ii) type $T_{1}: x_{2 n}$ can take any one of $2 n$ values, but its twin must be placed to separate the unique pair of neighboring twins in the new permutation. The recurrence formula follows:
$$F_{0}(n)=2 n\left[(2 n-2) F_{0}(n-1)+F_{1}(n-1)\right]$$
Now let $\left(x_{1}, \ldots, x_{2 n}\right)$ be a permutation of type $T_{1}$, and let $\left(x_{j}, x_{j+1}\right)$ be the unique neighboring twin pair. Similarly, on removing this pair we get a permutation of $2 n-2$ elements, either of type $T_{0}$ or of type $T_{1}$. The pair $\left(x_{j}, x_{j+1}\right)$ is chosen out of $n$ twin pairs and can be arranged in two ways. Also, in the first case it can be placed anywhere ( $2 n-1$ possible positions), but in the second case it must be placed to separate the unique pair of neighboring twins. Hence,
$$F_{1}(n)=2 n\left[(2 n-1) F_{0}(n-1)+F_{1}(n-1)\right]=F_{0}(n)+2 n F_{0}(n-1)$$
This implies that $F_{0}(n)2$, is the twin of $x_{1}$, then
$$f\left(x_{1}, x_{2}, \ldots, x_{2 n}\right)=\left(x_{2}, \ldots, x_{k-1}, x_{1}, x_{k}, \ldots, x_{2 n}\right)$$
The mapping $f$ is injective, but not surjective. Thus $F_{0}(n)<F_{1}(n)$.