$35 \cdot 13$ A child has a set of 96 different building blocks, each of which is either plastic or wooden, and comes in one of three sizes (large, medium, small), one of four colors (blue, green, red, yellow), and one of four shapes (circle, hexagon, square, triangle). The number of blocks in the set that differ from the "plastic, medium, red, circular block" in exactly two ways is
(A) 29 .
(B) 39 .
(C) 48 .
(D) 56 .
(E) 62 .
(40th American High School Mathematics Examination, 1989)
[Sol] The number of building blocks that differ in two aspects from the four given categories can be seen in the following table (note that there are 2 types of material, 3 types of size, 4 types of color, and 4 types of shape)
\begin{tabular}{c|c|c|c|c|c|c}
\hline Differences & Red, Round & Medium, Red & Medium, Round & Plastic, Medium & Plastic, Red & Plastic, Round \\
\hline Number of Types & $C_{3}^{1} C_{3}^{1}$ & $C_{2}^{1} C_{3}^{1}$ & $C_{2}^{1} C_{3}^{1}$ & $C_{1}^{1} C_{2}^{1}$ & $C_{1}^{1} C_{3}^{1}$ & $C_{1}^{1} C_{3}^{1}$ \\
\hline
\end{tabular}
(Here, the first character is used to represent each category)
Therefore, the total number is $3 \cdot 3 + 2 \cdot 3 + 2 \cdot 3 + 2 + 3 + 3 = 29$.
Hence, the answer is $(A)$.