The number of triangles whose vertices are at the vertices of a regular octagon but none of whose sides is a side of the octagon is

<p>To solve this problem, we need to determine the number of triangles formed by the vertices of a regular octagon such that none of the sides of the triangle is also a side of the octagon.</p> <p>Let's start by counting the total number of triangles that can be formed using the 8 vertices of the octagon. The number of ways to choose 3 vertices out of 8 is given by the combination formula:</p> <p> <p>$$ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 $$</p> </p> <p>Next, we need to exclude those triangles that have at least one side coinciding with a side of the octagon. Let's analyze how many such invalid triangles there can be.</p> <p>Consider each side of the octagon. For any given side, there are exactly 5 other vertices remaining (since we must exclude the two vertices that form the current side). Out of these 5 vertices, we can choose any 1 to form a triangle that has one side common with the octagon. Hence, for each side of the octagon, there are 5 such triangles.</p> <p>Since the octagon has 8 sides, the total number of triangles that have at least one side as a side of the octagon is:</p> <p> <p>$8 \times 5 = 40$</p> </p> <p>Therefore, the number of triangles whose sides do not coincide with any sides of the octagon is:</p> <p> <p>$56 - 40 = 16$</p> </p> <p>So, the correct answer is:</p> <p><b>Option B: 16</b></p>