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

Revision as of 18:54, 22 December 2015

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

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

Since no three points are collinear, every two points determine a distinct line. Thus, there are $\dbinom{12}{2} = \frac{12 * 11}{2} = 66$ lines.

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

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

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 == Solution ==
-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>
+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> lines.
 Therefore, the answer is <math>\fbox{(D) 66}</math>