1. Let $m, n \in \mathbf{N}$, and $m>n$, sets $A=\{1,2, \cdots, m\}, B=\{1,2, \cdots, n\}$,
and $C \subset A$. Then the number of $C$ such that $B \cap C \neq \varnothing$ is
$\qquad$ .
$=.1 .2^{m-n}\left(2^{n}-1\right)$
Set $A=B \bigcup\{n+1, n+2, \cdots, m\}$.
Since $B \cap C \neq \varnothing$, and $C \subset A$, thus, the elements of $C$ partly come from the non-empty subsets of $B$, and $B$ has $2^{n}-1$ non-empty subsets; the other part comes from the set $\{n+1, n+2, \cdots, m\}$, and the number of ways to select its elements is $2^{m-n}$.

According to the multiplication principle, the number of sets $C$ that satisfy the condition is $2^{m-n}\left(2^{n}-1\right)$.