$23 \cdot 4$ In a cube, among the 8 vertices, the midpoints of the 12 edges, the centers of the 6 faces, and the center of the cube, a total of 27 points, the number of groups of three collinear points is
(A) 57.
(B) 49.
(C) 43.
(D) 37.
(China High School Mathematics League, 1998)
[Solution] The number of collinear triplets with both endpoints being vertices is $\frac{8 \times 7}{2}=28$ (groups);
The number of collinear triplets with both endpoints being the centers of faces is $\frac{6 \times 1}{2}=3$ (groups);
The number of collinear triplets with both endpoints being the midpoints of edges is $\frac{12 \times 3}{2}=18$ (groups), and there are no other types of collinear triplets.

Therefore, the total number of collinear triplets among the given points is $28+3+18=49$. Hence, the answer is (B).