Asked by Nichole
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?
Answers
Answered by
MathMate
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 <i>over-counted</i> the number of triangles by 6, which is the number of ways to order three points.
The number of <i>distinct</i> 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
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 <i>over-counted</i> the number of triangles by 6, which is the number of ways to order three points.
The number of <i>distinct</i> 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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