Posted by **Nichole** on Wednesday, December 14, 2011 at 11:56pm.

There are nine points on a piece of paper. No three of the points are collinear. How many different triangles can be formed by using three of the nine points as vertices?

- math -
**MathMate**, Thursday, December 15, 2011 at 3:25am
Number of possible choices for the first point = 9.

Number of possible choices for the second point = 8.

Number of possible choices for the third point = 7.

Possible triangles with specific order of points = 9*8*7 = 504.

However, when we say triangle, we are not really concerned in which order the points are selected. So we have *over-counted* the number of triangles by 6, which is the number of ways to order three points.

The number of *distinct* triangles is therefore 504/6=84.

This number is mathematically called

9 choose 3, calculated by

9!/(3!(9-3)!) = 84

where 9! is factorial 9, = 9*8*7*...*2*1

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