2021 AMC 10A Problems/Problem 15: Difference between revisions
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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 14:00, 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 
.
