looks likeern5231 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.
2C4*5C1+5C2*4C1=30+40=70
what is oa?
looks likeern5231 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.
small correction in the calculation.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