Difference between revisions of "2021 AMC 10A Problems/Problem 15"
(added solution) |
(added formatting) |
||
Line 1: | Line 1: | ||
− | Assume that the first equation is above the second, since order doesn't matter. Then <math>C>A</math> and <math>B>D</math>. Therefore the number of ways to choose the four integers is <math>\tbinom{6}{2}\tbinom{4}{2}=90</math>. | + | ==Problem== |
+ | Values for <math>A,B,C,</math> and <math>D</math> are to be selected from <math>\{1, 2, 3, 4, 5, 6\}</math> without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves <math>y=Ax^2+B</math> and <math>y=Cx^2+D</math> intersect? (The order in which the curves are listed does not matter; for example, the choices <math>A=3, B=2, C=4, D=1</math> is considered the same as the choices <math>A=4, B=1, C=3, D=2.</math>) | ||
+ | |||
+ | <math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math> | ||
+ | |||
+ | ==Solution== | ||
+ | Assume that the first equation is above the second, since order doesn't matter. Then <math>C>A</math> and <math>B>D</math>. Therefore the number of ways to choose the four integers is <math>\tbinom{6}{2}\tbinom{4}{2}=90</math>, and the answer is <math>\boxed{C}</math>. | ||
+ | |||
+ | ==See also== | ||
+ | {{AMC10 box|year=2021|ab=A|num-b=14|num-a=16}} | ||
+ | {{MAA Notice}} |
Revision as of 15:02, 11 February 2021
Problem
Values for and are to be selected from without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves and intersect? (The order in which the curves are listed does not matter; for example, the choices is considered the same as the choices )
Solution
Assume that the first equation is above the second, since order doesn't matter. Then and . Therefore the number of ways to choose the four integers is , and the answer is .
See also
2021 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.