6. Let $A$ and $B$ be non-empty subsets of the set $\{1,2, \cdots, 10\}$, and the smallest element in set $A$ is not less than the largest element in set $B$. Then the number of such pairs $(A, B)$ is.
6. 9217.

When the smallest element in $A$ is $t(1 \leqslant t \leqslant 10)$, the sets $A$ and $B$ have $2^{10-t}$ and $2^{t}-1$ possibilities, respectively;

The pair of sets $(A, B)$ has $2^{10-t}\left(2^{t}-1\right)=2^{10}-2^{10-t}$ possibilities.
Therefore, the number of pairs satisfying the condition is
$$
\begin{array}{l}
\sum_{t=1}^{10}\left(2^{10}-2^{10-t}\right) \\
=10 \times 2^{10}-\left(2^{10}-1\right) \\
=9 \times 2^{10}+1=9217 \text { (pairs). }
\end{array}
$$