View source for 1952 AHSME Problems/Problem 31

== Problem ==

Given <math>12</math> points in a plane no three of which are collinear, the number of lines they determine is: 

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

== Solution ==

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>

== See also ==
{{AHSME 50p box|year=1952|num-b=30|num-a=32}}  

[[Category: Introductory Geometry Problems]]
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