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?
Option
a. 680
b. 720
c. 816
d. 640

• to form a triangle we need 3 points
  
  select 2 points from the 10 points of line AB & 1 from the 8 on AC
  = (10C2)*(8C1)
  
  select 2 points from the 8 points of line AC & 1 from the 10 on AB=
  (8C2)*(10C1)
  
  total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
  
  d.640
• 10 years agoHelpfull: Yes(97) No(5)
• 80 triangles also possible by taking point A as 3rd vertex
  
  so ans is 640+80=720
  
• 10 years agoHelpfull: Yes(11) No(15)
• To form a triangle we need three point...here we have total 18 points.
  
  So number of triangle will b 816.(18c3)
  
  But 10 on same line and other 8 are also on same line eventually they cant form triangles..
  
  So total number of triangle will b 18c3 -10c3 - 8c3 which is 816 -120 - 56 = 640 .
• 9 years agoHelpfull: Yes(6) No(1)
• Ans: (10 c1 * 8c1)+ (10 c2 *8 c 1) +(8 c 2* 10 c 1)=720
  
• 7 years agoHelpfull: Yes(5) No(0)
• Ans is 640 as (10C2)*(8C1)+ (8C2)*(10C1) = 640
  A triangle can be formed using 3 points so we can either chose 2 points from AB and 1 point from AC(10C2*8C1) or 1 point from AB and 2 points from AC(10C1*8C2).
  
• 9 years agoHelpfull: Yes(3) No(3)
• to form a triangle we need 3 points
  
  select 2 points from the 10 points of line AC & 1 of from the 8 on AB
  = (10C2)*(8C1)
  
  select 2 points from the 8 points of line AB & 1 of from the 10 on AC=
  (8C2)*(10C1)
  
  total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
  
  d.640
• 10 years agoHelpfull: Yes(2) No(4)
• to form a triangle we need 3 points
  
  select 2 points from the 10 points of line AB & 1 from the 8 on AC
  = (10C2)*(8C1)
  
  select 2 points from the 8 points of line AC & 1 from the 10 on AB=
  (8C2)*(10C1)
  
  total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
  
  d.640
• 9 years agoHelpfull: Yes(1) No(0)
• with vertice A there are 80 tringles possible, which are taken in to count with 640
• 8 years agoHelpfull: Yes(1) No(0)
• To form a triangle we need 3 vertices.So we can select one vertice from first straight line and two vertices from second straight line.In the same way we can have one more combination as shown below:
  (10c2*8c1)+(10c1*8c2)+(10c1*8c1)=720.
  The last combination (10c1*8c1) forms a triangle by taking single point from each straight line and thitd point as 'A'.
• 6 years agoHelpfull: Yes(1) No(0)
• RAKESH can u plz explain how are u getting 640? I mean the calculation.
• 9 years agoHelpfull: Yes(0) No(0)
• to form a triangle we need 3 points
  
  select 2 points from the 10 points of line AB & 1 from the 8 on AC
  = (10C2)*(8C1)
  
  select 2 points from the 8 points of line AC & 1 from the 10 on AB=
  (8C2)*(10C1)
  
  total no. of triangles = (10C2)*(8C1)+ (8C2)*(10C1) = 640
  
  d.640
• 9 years agoHelpfull: Yes(0) No(1)
• 10c2*8c1+10c1*8c2 =640
  as 3 points are required to form a triangle
• 9 years agoHelpfull: Yes(0) No(2)
• Atriangle can be formed by three points, this three point can be choose in this way
  10c1*8c2 +10c*8c1= 640
• 9 years agoHelpfull: Yes(0) No(0)
• 10C2*8C1+10C1*8C2=640
• 4 years agoHelpfull: Yes(0) No(0)
• 10!*9!*8!+8!*7!*9! = 640
• 1 year agoHelpfull: Yes(0) No(0)
• To form a traingle we need 3 points . Let two points from straight line with 10 points and one point from another 8 point line..and vice versa by applying combination.. 10C2*8C1 + 8C2*10C1
• 7 Months agoHelpfull: Yes(0) No(0)