joroivanov wrote:

I got

D.My way of solving the problem however is not very efficient. There's gotta be a faster way.

Basically we have this:

Line 1) --A-----B-----C--

Line 2) --D-----E-----F----G----H--

First, if A connects with any 2 other points from Line 2 there are exactly 10 possible triangles. Same for B and C. So far, 3x10 = 30 triangles

Then, if we "flip" the lines and have point D connect with any 2 other points from Line 1 exactly 3 triangles can be formed. Same for E, F, G and H. So, 5x3 = 15

Therefore, 15 + 30 = 45 triangles can be formed.

Nice approach!

Kind of a thinking-on-your-feet approach but very effective.

Here's the approach I was taught:

We need to select 3 points to form a triangle.

We can't get all 3 from the 5 marked in a line because that won't form a triangle. Same for the 3 marked in a line.

So we can select 2 from the 5 points and the third from the 3 points or vice versa. In quant that translates to

(5C2)(3C1) + (3C2)(5C1) = 45