Difference between revisions of "1952 AHSME Problems/Problem 31"
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− | Since no three points are collinear, every two points must determine a distinct line. Thus, there are <math>\dbinom{12}{2} = \frac{12 | + | Since no three points are collinear, every two points must determine a distinct line. Thus, there are <math>\dbinom{12}{2} = \frac{12\cdot11}{2} = 66</math> lines. |
Therefore, the answer is <math>\fbox{(D) 66}</math> | Therefore, the answer is <math>\fbox{(D) 66}</math> |
Latest revision as of 15:18, 28 February 2019
Problem
Given points in a plane no three of which are collinear, the number of lines they determine is:
Solution
Since no three points are collinear, every two points must determine a distinct line. Thus, there are lines.
Therefore, the answer is
See also
1952 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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All AHSME Problems and Solutions |
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