$1 \cdot 26$ If $X$ is a finite set, $|X|$ denotes the number of elements in set $X$, and $S, T$ are subsets of $\{1,2, \cdots, n\}$. If for every $s \in S, s > |T|$, and for every $t \in T, t > |S|$, then the ordered pair $(S, T)$ is called "admissible".

How many "admissible" ordered subset pairs are there for the set $\{1,2, \cdots, 10\}$? Prove your conclusion.
[Solution] Let $A_{n}$ be the number of "admissible" ordered subset pairs of $\{1,2, \cdots, n\}$. Suppose $(S, T)$ is an "admissible" ordered subset pair. Let $S$ have $i$ elements and $T$ have $j$ elements. By the problem, the $i$ elements of $S$ are chosen from $j+1, j+2, \cdots, n$. The $j$ elements of $T$ are chosen from $i+1, i+2, \cdots, n$. Therefore, we have
$$
A_{n}=\sum_{i=0}^{n} \sum_{j=0}^{n} C_{n-i}^{j} C_{n-j}^{i} .
$$

Since $i \leqslant n-j, j \leqslant n-i$, it follows that $\quad i+j \leqslant n$. Let $k=i+j$, then
$$
A_{n}=\sum_{i=0}^{n} \sum_{j=0}^{n-i} C_{n-i}^{j} C_{n-j}^{i}=\sum_{i=0}^{n} \sum_{k=i}^{n} C_{n-i}^{k-i} C_{n+i-k}^{i} .
$$

By exchanging the order of summation, we get
$$
A_{n}=\sum_{k=i}^{n} \sum_{i=0}^{n} C_{n+i-k}^{i} C_{n-i}^{k-i}=\sum_{k=0}^{n} C_{2 n-k+1}^{k} .
$$

The last sum in the above expression is the $(2n+2)$-th term $F_{2 n+2}$ of the Fibonacci sequence $\left(F_{0}=0, F_{1}=1, F_{n+2}=F_{n+1}+F_{n}\right)$.
Therefore, for this problem