2010 AMC 10B Problem 23

Let $a_{ij}$ be the entry in row $i$ and column $j.$ The increasing conditions force $a_{11}=1,$ $a_{33}=9,$ and $a_{22}$ to be $4,$ $5,$ or $6.$ If $a_{22}=4,$ then $\{a_{12},a_{21}\}=\{2,3\}.$ Choose which two of $5,$ $6,$ $7,$ and $8$ go in the bottom-left pair $\{a_{31},a_{32}\};$ the other two go in the top-right pair $\{a_{13},a_{23}\}.$ Each pair then has only one increasing order. There are $2\binom42=12$ arrays: two orders for $2,$ $3$ around the upper-left corner and $\binom42$ choices for the bottom-left pair. By reversing the digits and rotating the array, there are also $12$ arrays with center $6.$ If $a_{22}=5,$ choose the three entries in positions $a_{12},$ $a_{13},$ and $a_{23}.$ They can be any three of $\{2,3,4,6,7,8\}$ except $\{2,3,4\}$ or $\{6,7,8\};$ those two choices would put all three small or all three large entries on one side and violate a required comparison with the center. Every other choice uniquely determines the remaining entries and their increasing orders. This gives $\binom63-2=18$ arrays. The total is $12+18+12=42.$ Thus, D is the correct answer.