$15 \cdot 8$ for $\{1,2\} \subseteq x \subseteq\{1,2,3,4,5\}$ (where $x$ is a subset of $\{1,2,3,4,5\}$) the number of solutions is
(A) 2.
(B) 4.
(C) 6.
(D) 8.
(E) None of these.
(23rd American High School Mathematics Examination, 1972)
[Solution] The 8 different subsets of $\{1,2\}$ and $\{3,4,5\}$, when combined with each other, form the different solutions for $x$, and there are no other solutions. Specifically, the subsets $x$ are
$$
\begin{array}{l}
\{1,2\}, \quad\{1,2,3\}, \quad\{1,2,4\}, \quad\{1,2,5\}, \quad\{1,2,3,4\}, \\
\{1,2,3,5\}, \quad\{1,2,4,5\}, \quad\{1,2,3,4,5\} .
\end{array}
$$

Therefore, the answer is $(D)$.