Carefully Count Triangles

First, if the triangle involves the black base , then, as Geoff points out , we have to select 1 of 4 purple lines and 1 of 3 pink lines, so the number of such triangles is:

$4 \times 3 = 12$

Second, if the triangle doesn't involve the black base, then (at least) 2 of the lines must have the same color. Let's consider the case of 2 purple lines . For each pair of purple lines, we can cap it off with any of the 3 pink lines, so the number of such triangles is:

${ 4 \choose 2 } \times 3 = 18$

Third, we consider the case of 2 pink linkes . For each pair of pink lines, we can cap it off with any of the 4 purple lines, so the number of such triangles is

${ 3 \choose 2 } \times 4 = 12$

It is clear that there are no triangles which use 3 purple lines or 3 pink lines. Hence, in total, the number of triangles is $12 + 18 + 12 = 42$ .

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Is there a better way to approach this problem?

You can count it as follows.

For triangles having pink lines as a side and bottom left vertex of the large triangle as a vertex, we need to choose 2 points out of 5 for every of 3 pink lines: ${ 5 \choose 2 } \times 3 = 30$

For triangles having purple lines as a side and bottom right vertex of the large triangle as a vertex, we need to choose 2 points out of every $4-1=3$ points on each of 4 purple lines. We count $4-1$ as we bypass already counted triangles with bottom left vertex of the large triangle : ${ 3 \choose 2 } \times 4 = 12$

Total is $30+12=\boxed{42}$

It can be seen that every line intersects all other lines.So the the number of triangles will be the number of ways of choosing three lines from the 8 lines which is 8C3.But problems remain.It can be noticed that there are two vertices of the triangle ,in one of which 5 lines intersects each other and in the other vertice ,4 lines intersects each other.So no triangle will be formed if we choose three lines from that 5 lines or from that 4 lines.So by subtraction ,we get that the number of ways must be 8C3-(5C3+4C3)=42.

Had the triangle been larger, I'd probably have gone for combinatorics too, but this approach was very quick mentally:

There are 6 triangles involving (and below) the top vertex.

If we move one "unit" away from that vertex we can get to two vertices. For each of these vertices there are 5 (one less) triangles involving them that are below them.

Not including the two vertices at the base (which have no triangles below them), we have 12 vertices we can eventually move to.

So, continuing in this way, we get 6 + 5 + 5 + 4 + 4 + 4 + 3 + 3 + 3 + 2 + 2 + 1 = 42.