$15 \cdot 14$ Let the union of sets $A, B$ be $A \cup B=\left\{a_{1}, a_{2}, a_{3}\right\}$. When $A \neq B$, $(A, B)$ and $(B, A)$ are considered different pairs. Then the number of such $(A, B)$ pairs is
(A) 8 .
(B) 9 .
(C) 26 .
(D) 27 .
(China High School Mathematics League, 1993)
[Solution 1] Given $A \cup B=\left\{a_{1}, a_{2}, a_{3}\right\}$. Then, as subsets of this, $A, B$ can have at most 3 elements.
(1) If $A=\left\{a_{1}, a_{2}, a_{3}\right\}$, then the possible $B$ can be the empty set, a single-element set, a two-element set, or a three-element set. The number of such $B$ is
$$
C_{3}^{0}+C_{3}^{1}+C_{3}^{2}+C_{3}^{3}=2^{3} \text { sets. }
$$

At this time, $(A, B)$ has $C_{3}^{3} \cdot 2^{3}$ pairs.
(2) If $A$ is a two-element set $\left(\left\{a_{1}, a_{2}\right\}\right.$ or $\left\{a_{1}, a_{3}\right\}$ or $\left.\left\{a_{2}, a_{3}\right\}\right)$, then there are $C_{3}^{2}$

types, and the corresponding $B$ has $2^{2}$ sets $\left(C_{2}^{0}+C_{2}^{1}+C_{2}^{2}=2^{2}\right)$.
At this time, $(A, B)$ has $C_{3}^{2} \cdot 2^{2}$ pairs.
(3) If $A$ is a single-element set, then there are $C_{3}^{1}$ types, and the corresponding $B$ has 2 sets. At this time, $(A, B)$ has $C_{3}^{1} \cdot 2$ pairs.
(4) If $A$ is the empty set, then there is $C_{3}^{0}$ type, and the corresponding $B$ has one set. At this time, $(A, B)$ has $C_{3}^{0} \cdot 1$ pairs.
Therefore, the total number of such $(A, B)$ pairs is
$$
C_{3}^{3} \cdot 2^{3}+C_{3}^{2} \cdot 2^{2}+C_{3}^{1} \cdot 2+C_{3}^{0} \cdot 2^{0}=3^{3}=27 \text { pairs. }
$$

Thus, the answer is $(D)$.
[Solution 2] Let $P=\left\{a_{1}, a_{2}, a_{3}\right\}$, then the scenario where $A=B$ and $A \cup B=P$ has only one pairing: $A=P$ and $B=P$.

The number of pairings where $A \neq B$ must be even, so the total number of pairings must be odd. Therefore, options $(A)$ and $(C)$ can be ruled out.
A rough calculation shows that the number of pairings is greater than 9,
Thus, the answer is $(D)$.