# Difference between revisions of "2021 AMC 10A Problems/Problem 15"

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− | + | 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>. |

## Revision as of 15:00, 11 February 2021

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 .