Difference between revisions of "1952 AHSME Problems/Problem 31"
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== Solution == | == Solution == | ||
− | <math>\fbox{}</math> | + | |
+ | Since no three points are collinear, every two points determine a distinct line. Thus, there are <math>\dbinom{12}{2} = \frac{12 * 11}{2} = 66</math> | ||
+ | Therefore, the answer is <math>\fbox{(D) 66}</math> | ||
== See also == | == See also == |
Revision as of 19:53, 22 December 2015
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 determine a distinct line. Thus, there are
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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