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2021 AMC 10A Problems/Problem 15

Revision as of 14:00, 11 February 2021 by Icewolf10 (talk | contribs) (added solution)

Assume that the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$. Therefore the number of ways to choose the four integers is $\tbinom{6}{2}\tbinom{4}{2}=90$.

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