Difference between revisions of "1952 AHSME Problems/Problem 31"

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== Solution ==
 
== Solution ==
  
Since no three points are collinear, every two points must determine a distinct line. Thus, there are  <math>\dbinom{12}{2} = \frac{12 * 11}{2} = 66</math> lines.  
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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\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 $12$ points in a plane no three of which are collinear, the number of lines they determine is:

$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 54 \qquad \textbf{(C)}\ 120 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ \text{none of these}$

Solution

Since no three points are collinear, every two points must determine a distinct line. Thus, there are $\dbinom{12}{2} = \frac{12\cdot11}{2} = 66$ lines.

Therefore, the answer is $\fbox{(D) 66}$

See also

1952 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 30
Followed by
Problem 32
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All AHSME Problems and Solutions

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