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 Assume the four towns are at vertices of a square.

 There are  4C2 = 6 different roads, and so 6C3=20 ways to build 3 roads.

 A triangle of roads would need to have one road be a diagonal of the square, and for each of the 2 diagonals 2 triangles can be formed, so 4 triangles and number of ways to construct three roads each joining a pair of towns so that the roads do not form a triangle is 20 - 4 = 16 constructions satisfying the conditions.

Option D is the Correct Answer.

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