Given a set $C$ consisting of $n$ elements. In how many ways can two subsets $A$ and $B$ be chosen from $C$ such that

a) the sets $A$ and $B$ do not intersect;

b) the set $A$ is contained in the set $B$?
a) The problem is equivalent to partitioning the set $C$ into three subsets $A, B$, and $D=C \backslash(A \cup B)$. This, in turn, is equivalent to distributing $n$ different coins into three pockets (see problem $\underline{60348}$ ).
b) Instead of choosing subsets $A$ and $B$, one can choose non-intersecting subsets $A$ and $E$, and then set $B=A \cup E$. Therefore, parts a) and b) are equivalent.

## Answer

$3^{n}$ ways (in both cases).