If the sides AB, BC and CA of a â–³ABC have a, b and c points lying on them respectively, then the number of triangles that can be constructed using these points as vertices, if
A
two vertices lie on same side =a+b+cC3−(aC3+bC3+cC3+abc)
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B
two vertices lie on the same side =a+b+cC3−abc
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C
all the vertices lie on different sides =abc
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D
all the vertices lie on different sides =a+b+cC3−(aC3+bC3+cC3)
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Solution
The correct option is C all the vertices lie on different sides =abc Total number of triangles = total number of ways of choosing 3 points − number of ways of choosing all the 3 points from AB or BC or CA =a+b+cC3−(aC3+bC3+cC3)
Number of triangle formed such that all the vertices lies on different sides =aC1⋅bC1⋅cC1=abc
∴ Number of triangles formed such that two of the vertices lies on same side =a+b+cC3−(aC3+bC3+cC3+abc)