Problem 1336

To solve the problem, we need to count the number of admissible ordered pairs $(S, T)$ of subsets of $\{1, 2, \cdots, 10\}$ such that $s > |T|$ for each $s \in S$ and $t > |S|$ for each $t \in T$.

1. Understanding the conditions:
- For each element $s \in S$, $s > |T|$.
- For each element $t \in T$, $t > |S|$.

2. Analyzing the constraints:
- If $S$ is non-empty, let $m = |S|$. Then, for each $t \in T$, $t > m$. This means $T$ can only contain elements greater than $m$.
- Similarly, if $T$ is non-empty, let $n = |T|$. Then, for each $s \in S$, $s > n$. This means $S$ can only contain elements greater than $n$.

3. **Considering the possible values for $S$ and $T$**:
- If $S$ and $T$ are both non-empty, the smallest element in $S$ must be greater than the size of $T$, and the smallest element in $T$ must be greater than the size of $S$. This creates a contradiction unless both sets are empty.

4. Counting the admissible pairs:
- The only way to satisfy both conditions without contradiction is if both $S$ and $T$ are empty. Therefore, the only admissible pair is $(\emptyset, \emptyset)$.

5. Conclusion:
- There is exactly one admissible ordered pair of subsets of $\{1, 2, \cdots, 10\}$.

The final answer is $\boxed{1}$.