(2) Let $A$ and $B$ be two sets, and call $(A, B)$ a "pair". When $A \neq B$, consider $(A, B)$ and $(B, A)$ as different "pairs". The number of different pairs $(A, B)$ that satisfy the condition $A \cup B = \{1,2,3,4\}$ is $\qquad$.
(2) 81 Hint: When set $A$ has no elements, i.e., $A=\varnothing$, set $B$ has 4 elements, there is 1 case; when set $A$ contains 1 element, set $B$ contains the other 3 elements besides this element, and the element in set $A$ may or may not be in set $B$, there are $\mathrm{C}_{4}^{1} \times 2=8$ cases; when set $A$ contains 2 elements, set $B$ contains the other 2 elements besides these 2 elements, and the elements in set $A$ may or may not be in set $B$, there are $\mathrm{C}_{4}^{2} \times 2^{2}=24$ cases; when set $A$ contains 3 elements, set $B$ contains the other 1 element besides these 3 elements, and the 3 elements in set $A$ may or may not be in set $B$, there are $C_{4}^{3} \times 2^{3}=32$ cases; when set $A$ contains 4 elements, the elements in set $B$ may or may not be in set $A$, there are $\mathrm{C}_{4}^{4} \times 2^{4}=16$ cases. In summary, the number of "pairs" is
$$
1+8+24+32+16=81 \text {. }
$$