Difference between revisions of "1952 AHSME Problems/Problem 31"
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== Solution == | == Solution == | ||
| − | Since no three points are collinear, every two points 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 14:18, 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 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
