1952 AHSME Problems/Problem 31

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 determine a distinct line. Thus, there are $\dbinom{12}{2} = \frac{12 * 11}{2} = 66$ Therefore, the answer is $\fbox{(D) 66}$

1952 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 30	Followed by Problem 32
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1952 AHSME Problems/Problem 31

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