There are 12 points in a plane, no three of which are in the same straight line, except 5 …

Question

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

Accepted Answer

To count the number of distinct triangles:

Start with all possible triples of points.

From 12 points, the number of ways to choose any 3 is

$      \binom{12}{3}= \frac{12\cdot11\cdot10}{3\cdot2\cdot1}=220.    $

Subtract the ``invalid'' triples that are collinear.

The only collinear sets of three points arise from the single line containing the 5 collinear points.

Number of ways to pick 3 points from those 5 is

$      \binom{5}{3}=10.    $

Valid triangles = total triples - collinear triples.

$      220-10 = 210.    $

Hence, the total number of triangles that can be formed is 210.

Correct option: B

Solutions · 1