Difference between revisions of "2021 AMC 10A Problems/Problem 15"
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==Solution== | ==Solution== | ||
| − | + | Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now 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>. ~IceWolf10 | |
==See also== | ==See also== | ||
{{AMC10 box|year=2021|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2021|ab=A|num-b=14|num-a=16}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 14:04, 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
Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now 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 
. ~IceWolf10
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. 
