If 4 points are indicated on a line and 5 points are indicated on another line that is parallel to the first line, how many triangles can be formed whose vertices are among the 9 points?
a)20
b)30
c)40
d)70
e)90
answer is D.
I immediately recognized that it was a combinatronics problem but couldn't seem to rationalize a formula in my head. I initially tried drawing the 2 lines to see what kind of relationship I can get. I chose the top to be the 5 pointed line and the bottom to be 4 pointed line. I decided draw the triangle from top to botto m (1 point top 2 points bottom) I came to the formula that for every 1 point on top there will be 4!/2!2! on bottom so 6...and for total # triangles on top 6X5=30 (triangles possible w 1 point on top 2 points bottom) and then flipped it to do the same thing 5!/2!3!=10 x 4=40 30+40=70. Even though this makes sense to me now it took me forever to rationalize it. Is there something that I am not seeing that should trigger a quicker response? Thanks!