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The sides AB, BC, CA of a triangle ABC have 3, 4 and 5 interior points respectively on them. The number of triangles that can be constructed using these points as vertices is

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answer is 205.

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Detailed Solution

We have, in all 12 points. Since 3 points are used to form a triangle, therefore the total number of triangles,

including the triangles formed by collinear points on $A B, B C$ and $C A$ , is $^{12} C_{3} = 220 .$

But, this includes the following:

The number of triangles formed by 3 points on $A B$

$=^{3} C_{3} = 1$ ,

The number of triangles formed by 4 points $B C$

$=^{4} C_{3} = 4,$

The number of triangles formed by 5 points on $C A$

$=^{5} C_{3} = 10,$

Hence, p· :• ,:, number of triangles $= 220 - (10 + 4 + 1) = 205 .$

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