The correct answer is 20.

The three-dimensional figure shown in the question is a regular octahedron — a symmetric 3D solid with 6 vertices, 12 edges, and 8 triangular faces. The key insight for this problem is that we must count all possible triangles that can be formed using the vertices of this figure, not just the visible faces.

Step 1: Identify the number of vertices
A regular octahedron has exactly 6 vertices. Label them as V₁, V₂, V₃, V₄, V₅, and V₆.

Step 2: Understand what forms a triangle
A triangle is formed by choosing any 3 non-collinear points. In a regular octahedron, no three vertices are collinear (no three vertices lie on the same straight line). This is a critical geometric property of the octahedron. Therefore, every possible selection of 3 vertices from the 6 available vertices will always form a valid triangle.

Step 3: Apply the Combination Formula
The number of ways to choose 3 vertices from 6 is given by the combination formula:

$${}^{}{C}_{}(n,r)=\frac{n!}{r!(n-r)!}$$

Here, n = 6 (total vertices) and r = 3 (vertices needed to form a triangle):

$$C(6,3)=\frac{6!}{3!(6-3)!}=\frac{6!}{3!3!}$$

Step 4: Simplify the calculation

$$=\frac{6\times 5\times 4\times 3!}{3!\times 3!}=\frac{6\times 5\times 4}{3\times 2\times 1}=\frac{120}{6}=20$$

Step 5: Conclusion
Since no three vertices of the octahedron are collinear, all 20 combinations of 3 vertices yield a valid triangle. These include the 8 outer faces of the octahedron itself plus 12 additional internal/diagonal triangles formed by connecting non-adjacent vertices through the interior of the figure.

Therefore, the total number of triangles in the three-dimensional figure is 20.