Let A , B , and C be the vertices of a triangle. There are 3 points on the side AB (excluding A and B ), 4 points on the side BC (excluding B and C ), and 5 points on the side CA (excluding C and A ). The total number of triangles that can be formed using these 12 points along with the 3 vertices A , B , and C as vertices is:

Step-by-step solution

Total number of points available is 3 (on AB ) + 4 (on BC ) + 5 (on CA ) + 3 (vertices A, B, C ) = 15 points. The total number of ways to choose any 3 points from these 15 points is ¹⁵C₃ . However, any 3 points chosen from the same side of the triangle will be collinear and will not form a triangle. We must subtract these cases. The number of points on the line segment AB (including vertices A and B ) is 3 + 2 = 5 . The number of points on the line segment BC (including vertices B and C ) is 4 + 2 = 6 . The number