Answer: 96
Step-by-step explanation:
Ok, lines a and b are parallel.
We can separate this problem in two cases:
Case 1: 2 vertex in line a, and one vertex in line b.
Here we use the relation:
"In a group of N elements, the total combinations of sets of K elements is given by"
[tex]C = \frac{N!}{(N - K)!*K!}[/tex]
Here, the total number of points in the line is N, and K is the ones that we select to make the vertices of the triangle.
Then if we have two vertices in line a, we have:
N = 6, K = 2
[tex]C = \frac{6!}{4!*2!} = \frac{6*5}{2} = 3*5 = 15[/tex]
And the other vertex can be on any of the four points on the line b, so the total number of triangles is:
C = 15*4 = 60.
But we still have the case 2, where we have 2 vertices on line b, and one on line a.
First, the combination for the two vertices in line b is:
We use N = 4 and K = 2.
[tex]C = \frac{4!}{2!*2!} = \frac{4*3}{2} = 6[/tex]
And the other vertice of the triangle can be on any of the 6 points in line a, so the total number of triangles that we can make in this case is:
C = 6*6 = 36
Then, putting together the two cases, we have a total of:
60 + 36 = 96 different triangles
