ern5231 wrote:Consider two parallel lines(one below other).Above line has 5 points and the line below has 4 points. How many triangles are possible with these points?

I may post the image later.

looks like
2C4*5C1+5C2*4C1=30+40=70

what is oa?

total number of triangle formed with 5+4 point =9C3=84
total number of triangle formed with 5 point =4C3=10
total number of triangle formed with 4 point =4C3=4

Since points in the same line will not form triangle,total number of triangles is
[spoiler]84-10-4= 70[/spoiler]

Should the solution not be :

C (5,2) xC (4,1) -> Selecting two points from above line and one from below

C (5,1) xC (4, 2) -> Selecting one point from above line and two from below

C (5,2) xC (4,1) + C (5,1) xC (4, 2) = 50

I think the mistake in the posts above is consideration of 2C4 instead of 4C2. I not quite good with Combinatorics so pls correct me if I am wrong

ern5231 wrote:Should the solution not be :

C (5,2) xC (4,1) -> Selecting two points from above line and one from below =5*4*4/2=40

C (5,1) xC (4, 2) -> Selecting one point from above line and two from below =5*4*3/2=30

C (5,2) xC (4,1) + C (5,1) xC (4, 2) =40+30= 50

I think the mistake in the posts above is consideration of 2C4 instead of 4C2. I not quite good with Combinatorics so pls correct me if I am wrong

small correction in the calculation.