On a triangle ABC, on the side AB, 5 points are marked, 6 points are marked on side BC, and 3 points are marked on the AC ( none of the points being the vertex of the triangle). How many triangles can be made using these points.
option
a) None of these
b) 328
c) 333
d) 90

• Total there are 5+6+3=14 points.... so to form a triangle we need 3 sides...that means 14c3 ways... and on each side there are 5,6,3 points ...as they are collinear no triangle can be drawn out of them...thus it is 14c3-(5c3+6c3+3c3)=333
• 10 years agoHelpfull: Yes(63) No(0)
• required no of triangles=5c1*6c1*3c1+(5c2*6c1+5c2*3c1)+(6c2*5c1+6c2*3c1)+(3c2*5c1+3c2*6c1)=90+90+120+33=333 triangles
• 10 years agoHelpfull: Yes(9) No(0)
• There are two possibilities.
  
  Case 1. One vertex on each side.
  There are 5*6*3=90 possibilities
  
  Case 2. Two vertices on the same side.
  We have
  5C2*(6+3)+6C2*(5+3)+3C2*(5+6)= 243
  possibilities.
  
  Thus in total, there are 90+243=333 possible triangles.
• 5 years agoHelpfull: Yes(1) No(0)
• MANISH U ARE WRONG
  
• 10 years agoHelpfull: Yes(0) No(20)
• may be....
  14c3=364
  
• 10 years agoHelpfull: Yes(0) No(8)
• total number of points are 14(5+6+3). to form a triangle we need 3 points, so number of triangles using these points would be 14c3 -(5c3+6c3+3c3)=333
• 5 years agoHelpfull: Yes(0) No(0)