AMC10 2010 B

Answer (D): Let $a_{ij}$ denote the entry in row $i$ and column $j$. The given conditions imply that $a_{11}=1$, $a_{33}=9$, and $a_{22}=4,5,$ or $6$. If $a_{22}=4$, then $\{a_{12},a_{21}\}=\{2,3\}$, and the sets $\{a_{31},a_{32}\}$ and $\{a_{13},a_{23}\}$ are complementary subsets of $\{5,6,7,8\}$. There are $\binom{4}{2}=6$ ways to choose $\{a_{31},a_{32}\}$ and $\{a_{13},a_{23}\}$, and only one way to order the entries. There are $2$ ways to order $\{a_{12},a_{21}\}$, so $12$ arrays with $a_{22}=4$ meet the given conditions. Similarly, the conditions are met by $12$ arrays with $a_{22}=6$. If $a_{22}=5$, then $\{a_{12},a_{13},a_{23}\}$ and $\{a_{21},a_{31},a_{32}\}$ are complementary subsets of $\{2,3,4,6,7,8\}$ subject to the conditions $a_{12}<5$, $a_{21}<5$, $a_{32}>5$, and $a_{23}>5$. Thus $\{a_{12},a_{13},a_{23}\}\ne\{2,3,4\}$ or $\{6,7,8\}$, so its elements can be chosen in $\binom{6}{3}-2=18$ ways. Both the remaining entries and the ordering of all entries are then determined, so $18$ arrays with $a_{22}=5$ meet the given conditions. Altogether, the conditions are met by $12+12+18=42$ arrays.