# 1992 AHSME Problems/Problem 22

## Problem

Ten points are selected on the positive $x$-axis, $X^+$, and five points are selected on the positive $y$-axis, $Y^+$. The fifty segments connecting the ten points on $X^+$ to the five points on $Y^+$ are drawn. What is the maximum possible number of points of intersection of these fifty segments that could lie in the interior of the first quadrant? $\text{(A) } 250\quad \text{(B) } 450\quad \text{(C) } 500\quad \text{(D) } 1250\quad \text{(E) } 2500$

## Solution $\fbox{B}$ We can pick any two points on the $x$-axis and any two points on the $y$-axis to form a quadrilateral, and the intersection of its diagonals will thus definitely be inside the first quadrant. Hence the answer is $10C2 \times 5C2 = 450$.

## See also

 1992 AHSME (Problems • Answer Key • Resources) Preceded byProblem 21 Followed byProblem 23 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 All AHSME Problems and Solutions

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