10 points on straight line on ab and 8points on straight line ac then how many number of triangles are formed
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Answer:
504
Step-by-step explanation:
let Point - A is common on both lines .
total no of ∆ = ∆ of type-1 + ∆ of type-2 + ∆ of type-3
= 63 + 189 + 252 = 504
∆ of type-1 = A as one vertex and other two vertices on different lines
= 1 × 9C1 × 7C1
= 63
now, excluding point A :
∆ of type-2 = one vertex on AB and other two vertices on AC
= 9C1 × 7C2
= 9 × 7×6/2
= 189
∆ of type-3 = one vertex on AC and other two vertices on AB
= 7C1 × 9C2
= 7 × 9×8/2
= 252
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