How many distinct triangles can be constructed by connecting three different vertices of a cube? (Two triangles are distinct if they have different locations in space.)
To solve this problem, I need to count the number of ways to choose three vertices of a cube out of the eight available. This is a combination problem, since the order of the vertices does not matter. The formula for combinations is nCr = n! / (r! (n-r)!), where n is the total number of elements and r is the number of elements chosen. In this case, n = 8 and r = 3, so nCr = 8! / (3! (8-3)!) = 56. Therefore, there are 56 distinct triangles that can be constructed by connecting three different vertices of a cube.

