2010 AMC 10B Problem 23

Write the cell in row $i$, column $j$ as $(i,j)$. Clearly $1$ is at $(1,1)$ and $9$ at $(3,3)$. The center $(2,2)$ exceeds $(1,1), (1,2), (2,1)$ and is below $(2,3), (3,2), (3,3)$, so the center is $4$, $5$, or $6$.

Center $4$. Then $\{2, 3\}$ fill $(1,2)$ and $(2,1)$ in $2$ ways. The digits $5, 6, 7, 8$ go to $(1,3), (2,3), (3,1), (3,2)$ with only the constraints $(1,3) < (2,3)$ and $(3,1) < (3,2)$: choose which two digits go to the top-right pair, $\binom42 = 6$ ways. Total $2 \cdot 6 = 12$.

Center $6$. Replacing each entry $x$ by $10 - x$ and rotating $180^\circ$ swaps this case with center $4$: also $12$.

Center $5$. Three of $\{2,3,4\}$ must fill $(1,2), (2,1)$ and one of the corners $(1,3)$ or $(3,1)$; the other corner gets a digit from $\{6,7,8\}$. Say $(1,3) < 5$: then $(2,1)$ can be any of $2, 3, 4$ ($3$ ways) and the other two fill row $1$ in increasing order. The digits $6, 7, 8$ go to $(2,3), (3,1), (3,2)$ with $(3,1) < (3,2)$: choose the digit for $(2,3)$, $3$ ways. That is $9$ arrays; by the transpose symmetry, $(3,1) < 5$ gives $9$ more. Total $18$.

Altogether $12 + 18 + 12 = 42$.

The answer is $\boxed{\textbf{(D)}\ 42}$.

Ordering constraints on a grid are best attacked at the most constrained cell, the center, which splits the digits into "before" and "after" sets. Each case then decomposes into small independent chains. Symmetries (transpose, complement $10 - x$) halve the work and act as a check on the counts.

These arrays are standard Young tableaux of a $3\times 3$ square, counted by the hook length formula: $\dfrac{9!}{5\cdot4\cdot3\cdot4\cdot3\cdot2\cdot3\cdot2\cdot1} = \dfrac{362880}{8640} = 42$.

• Assuming the center must be 5, or forgetting that entries in the two off-diagonal corners are not compared with each other.
• Overcounting the center-$5$ case by allowing both corners $(1,3)$ and $(3,1)$ to be less than $5$, which would need four digits below $5$ but only three exist.