There are 10 points on a straight line AB and 8 on another straight line AC
none of them being point A. how many triangles can be formed with these points
as vertices?

• 10C2*8C1+8C2*10C1
  =640
• 6 years agoHelpfull: Yes(14) No(3)
• Ans: 720
  You can choose any two points on AB and one point on AC : (10C2)*(8C1)
  Similarly , you can choose any two points on AC and one point on AB: (8C2)*(10C1)
  and Since two lines intersect at a you can join two points each on AB and AC with a line to form a triangle , therefore - (8 * 10)
  
  so , (10C2)*(8C1) + (8C2)*(10C1) + 80 = 720
• 6 years agoHelpfull: Yes(8) No(0)
• My answer assumes that you can't use A, B or C but only the other un-named points.
  
  Use any 2 from the 10 on line AB with 1 of those from the 8 on AC = (10C2)*(8C1)
  
  Use any 2 from the 8 on line AC with 1 of those from the 10 on AB = (8C2)*(10C1)
• 6 years agoHelpfull: Yes(3) No(0)
• 10c2*8c1+8c2*10c1=360+280=640
• 6 years agoHelpfull: Yes(1) No(0)
• How?? Plz explain
• 6 years agoHelpfull: Yes(0) No(0)
• =10c2*8c1+10c1*8c2
  =360+280
  =540
• 6 years agoHelpfull: Yes(0) No(6)
• We have 19 points total. We need to choose 3 of them. So the formula is 19C3=969 triangles
• 6 years agoHelpfull: Yes(0) No(1)
• Here given condition is A is excluded,so the formula to find the no.of triangles is
  No.of triangles=mn(m+n)/2
  m=10
  n=8
  No of triangles=10*8(10+8)/2
  =180*9
  = 720
• 3 years agoHelpfull: Yes(0) No(0)