AB is a straight line and 10 points are drawn in it. AC another straight 8 points are drwan in it. how many triangle can be made with these points as a vertices?

• 8c2*10c1 +8c1*10c2=640 ways
• 11 years agoHelpfull: Yes(20) No(1)
• 9c2*7c1 + 7C2*9c1 +7C1*9C1,,according to me i thnk we hv to consider triangles formd by common point A
• 11 years agoHelpfull: Yes(8) No(1)
• In AB line,10 points and in AC 8 points..
  So,to form a triangle we need 3 points..
  Thus,We can do it in (10c2*8c1)*(8c2*10c1) ways....
• 11 years agoHelpfull: Yes(5) No(8)
• We can think in two ways.
  1. 18c3-10c3-8c3=640
  2. 10c2*8c1 + 8c2*10c1=640
• 11 years agoHelpfull: Yes(5) No(0)
• Sorry,my given answer is wrong..It will be (10c2*8c1)+(8c2*10c1)..Sorry for the mistake ...
• 11 years agoHelpfull: Yes(1) No(1)
• 8c2*10c1 +8c1*10c2=640 ways
• 11 years agoHelpfull: Yes(1) No(0)
• 9c1*7c2 + 7c1*9c2 + 9c1*7c1
• 11 years agoHelpfull: Yes(1) No(0)
• I think A point will be common to both straight lines. So i think "9c1*7c1 + 10c2*7c1 + 10c1*7c2 + 8c1*9c2 + 8c2*9c1 = 1128 ways " will be the ans ...
• 11 years agoHelpfull: Yes(1) No(0)
• ans is 539
• 11 years agoHelpfull: Yes(0) No(3)
• sry it will be 594
• 11 years agoHelpfull: Yes(0) No(3)
• its 7c1*10c2+8c2*9c1
• 11 years agoHelpfull: Yes(0) No(0)
• 8c2*10c1 +8c1*10c2-7(because a first point a and 7 point of ac can n't make triangle with a)-9( because a first point a and 9 point of ab can n't make triangle with a)=624
• 11 years agoHelpfull: Yes(0) No(1)
• 624
  8c2*10c1
  
  
• 11 years agoHelpfull: Yes(0) No(0)
• ANS: 640 ways
  To form a triangle, we need 3 points
  there are 2 ways of it:
  1 is to take 2 points from points on AB and 1 point each from AC OR vice verse
  = (10C2*8C1)+(10C1*8C2)= 640 WAYS
• 11 years agoHelpfull: Yes(0) No(0)
• 1 point is common to both..
  so ans will be
  10c2*7c1+7c2*10c1+7c1*9c1
• 11 years agoHelpfull: Yes(0) No(0)