6. Let $n>1$ be a positive integer, choose two non-empty subsets $A$ and $B$ from the set $S=\{1,2,3, \cdots, n\}$, such that the smallest number in $B$ is greater than the largest number in $A$. Then the number of different ways to choose is $\qquad$ kinds.
6. $n \cdot 2^{n-1}+1$ Detailed Explanation: Let the largest number in $A$ be $k$, then $A=A^{\prime} \cup\{k\}$, and $A^{\prime} \subseteq\{1,2, \cdots, k-1\}$, while $B \subseteq\{k+1, k+2, \cdots, n\}$, thus the number of non-empty subset pairs $(A, B)$ is $2^{k-1}\left(2^{n-k}-1\right)=2^{n-1}-2^{k-1}$; by the principle of classification counting, the total number of different selection methods is $\sum_{k=1}^{n-1}\left(2^{n-1}-2^{k-1}\right)=n \cdot 2^{n-1}+1$.