How many triangles can be formed when each vertex of the triangle is

The reason your answer is slightly too large is that you're counting some triangles twice - in your first case, you're counting triangles that use the middle dot, along with one dot on the horizontal line and one on the vertical line. But in your second case, you're counting those triangles again. There are 80 such triangles (as demonstrated above in Fdambro294's post), and because you counted each of them twice, your answer is too big by exactly 80.

Because that middle dot leads to complications (various cases) when you try to count the triangles directly, I solved this problem the same way gmatinsight did above: we have 19C3 ways to pick any three dots, and when we do that, we always get a triangle except when we pick three dots along the same line. So if we subtract 11C3 (the number of ways to pick three dots from the horizontal line) and 9C3 (the number of ways to pick 3 dots from the vertical line), we'll get the answer.

Any 3 points - 3 points in a line

19C3 - 11C3 - 9C3 = 969 - 165 - 84 = 720

Answer: Option B

1st Concept: For any Side of a Triangle, 2 of the Vertices must be in a Straight Line (here we have 2 Straight Lines: Horizontal and Vertical)

So I 1st thought picking 2 from the Horizontal and a 3rd Vertex from the Vertical would be the way to go. However, the middle dot caused issues.

So the attempt is to split it up into 3 different cases. For the 1st 2 Cases we're going to IGNORE the Middle Dot.

Case 1: Choosing 2 Vertices on the Horizontal Line and the 3rd Vertex on the Vertical Line to make a Triangle.

Ignoring the Middle dot, the No. of different ways we can choose 2 Different Vertices on the Horizontal Line = 10 - C - 2 = 45 Different Combinations

and

For Each Combination, we can pick 1 of the 8 Dots (again ignoring the Middle Dot) on the Vertical Line for the 3rd Vertex.

45 * 8 = 360 Different Triangles


Case 2: Choosing 2 Vertices on the Vertical Line and the 3rd Vertex on the Horizontal Line to make a Triangle

on the Vertical Line, we have 8 unique dots of which we can choose 2. The Total No. of Different Ways to choose 2 Dots out of 8 = 8 - C - 2 = 28

and

For each of those 28 Different combinations of 2 Vertices, we have 10 dots on the Horizontal Line that can make up the 3rd Vertex (again, ignoring the Middle dot).

28 * 10 = 280 Different Triangles


Case 3: Using the Middle Dot as 1 of the Vertices of the Triangle

The Key in this case is that if we use the Middle Dot as 1 of the Vertices of the Triangle, then we have to Choose 1 Vertex on the Horizontal line AND then Choose 1 Vertex on the Vertical Line.

So for Each Triangle that uses the Middle Dot as a Vertex:

1st: There is 10 options on the Horizontal Line from which to choose to make 1 Side

and

2nd: There is 8 options on the Vertical Line from which to choose to connect the 3rd Side

10 * 8 = 80 Possible Triangles


SUM up the 3 Cases

360 + 280 + 80 = 720 Possible Triangles that can be formed.

Answer Choice B

IMO
Two dots from Horizontal and One dot from vertical. ( Notice we cannot use the dot in the centre.
11C2*8 = 440
Two dots from Vertical and One dot from Horiz.
9c2*10 = 360

D = 800

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