There are seven points on a piece of paper. Exactly four of these points are on a straight line. No other line contains more than two of these points. Three of these seven points are selected to form the vertices of a triangle. How many triangles are possible?
(A) 18
(B) 28
(C) 30
(D) 31
(E) 33
There are seven points on a piece of paper. Exactly four of these points are on a straight line. No other line contains more than two of these points. Three of these seven points are selected to form the vertices of a triangle. How many triangles are possible?
(A) 18
(B) 28
(C) 30
(D) 31
(E) 33

## Solution

There are three cases to consider corresponding to zero vertices, one vertex or two vertices on the given line.

Case 1 'zero vertices on the given line'

Since there are exactly three points not on the line, there can only be one triangle formed with these three points.

Case 2 'one vertex on the given line'

There are four choices for the point on the line and for each of these four points there are three ways of selecting the pair of vertices not on the line. Thus, there are $3 \times 4$ or 12 possible triangles.

Case 3 ' 2 vertices on the given line'

There are six ways of choosing the pair of points on the line and for each of these six pairs there are three ways of selecting the vertex not on the line giving a total of $6 \times 3$ or 18 possibilities.

In total there are $1+12+18$ or 31 triangles.

ANSWER: (D)