The following figure shows a cube.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-10.jpg?height=453&width=497&top_left_y=1624&top_left_x=666)

a) Calculate the number of triangles whose three vertices are vertices of the cube.

b) How many of these triangles are not contained in a face of the cube?
Solution

a) Let's start by fixing a vertex, say $A$, and counting the number of triangles that use this vertex. Each of these triangles will be determined by indicating its other two vertices distinct from $A$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-32.jpg?height=294&width=349&top_left_y=556&top_left_x=904)

Then the number of triangles that use the vertex $A$ coincides with the number of ways to choose two distinct vertices from the set $\mathcal{V}=\{B, C, D, E, F, G, H\}$. Once the first vertex is chosen, there are 6 possible choices in $\mathcal{V}$ for the second vertex. We would thus count 7 (the number of vertices in $\mathcal{V}$) times 6 pairs of vertices in $\mathcal{V}$, but each pair would have been counted twice. Therefore, there are $7 \times 6 / 2 = 21$ ways to choose two distinct vertices in $\mathcal{V}$.

So far, we have proven that for each vertex of the cube, the number of triangles that use that vertex is 21. Since the cube has 8 vertices, we would count $8 \times 21$ triangles, but each triangle would have been counted three times, once for each of its vertices. We conclude that there are

$$
\frac{8 \times 21}{3} = 56
$$

triangles that can be formed using the vertices of the cube.

b) It is a good idea to first count the number of triangles that are contained in some face. Let's focus on a face, say the face $A B C D$.

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-32.jpg?height=344&width=389&top_left_y=1813&top_left_x=905)

To determine each triangle, we must choose its 3 vertices from the set of vertices $\{A, B, C, D\}$. Choosing a trio in $\{A, B, C, D\}$ is equivalent to choosing an element in the same set, the one that does not belong to the trio. There are then 4 trios, and therefore 4 triangles that are contained in the face $A B C D$. Since the cube has 6 faces, we would thus count $6 \times 4 = 24$ triangles. In this count, each triangle was counted exactly once, so the total number of triangles that are contained in the faces of the cube is 24. Thus, there are $56 - 24 = 32$ triangles that are not contained in a face of the cube.