A triangle ABC has 2 points marked on side BC, 5 points marked on side CA and 3 points marked on side AB. None of these marked points is coincident with the vertices of the triangle ABC. All possible triangles are constructed taking any three of these points and the points A, B, C as the vertices. How many new triangles have at least one vertex common with the triangle ABC?

1) 126
2) 129
3) 127
4) 128

• Answer is 127. Take vertex A as the only point on new triangle. We need to choose 2 points from the remaining 10 points which is 10C2=45. But we cannot choose 3 points in the same line so 3C2-5C2. Similar for other vertices and we get a total of 45x3 - 2(5C2+3C2+2C2) = 107. Now taking both A and B on the new trangle we can choose 5+2=7 triangles. similar for other vertices and we get more 2(5+2+3) = 20 triangles. So a total of 127 triangles.
• 4 years agoHelpfull: Yes(0) No(1)