2010 AMC 12B Problem 17

Write $a_{ij}$ for the entry in row $i,$ column $j.$ The conditions force $a_{11}=1,$ $a_{33}=9,$ and $a_{22}\in\{4,5,6\}.$ If $a_{22}=4,$ then $\{a_{12},a_{21}\}=\{2,3\}$ and $\{5,6,7,8\}$ split as complementary pairs filling the rest of the last row and column: $\binom42=6$ splits times $2$ orders for $\{2,3\}$ gives $12$ arrays. By symmetry $a_{22}=6$ also gives $12.$ 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 ordering constraints. The first set can be any three-element subset except $\{2,3,4\}$ or $\{6,7,8\},$ giving $\binom63-2=18$ arrays. Altogether $12+12+18=42.$ Thus, the correct answer is D.