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 overcounted 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!(93)!) = 84
where 9! is factorial 9, = 9*8*7*...*2*1
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