# Difference between revisions of "2010 AMC 12B Problems/Problem 17"

## Problem

The entries in a $3 \times 3$ array include all the digits from $1$ through $9$, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?

$\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60$

## Solution

The first 4 numbers will form one of 3 tetris "shapes".

First, let's look at the numbers that form a 2x2 block, sometimes called tetris $O$:

$\begin{tabular}{|c|c|c|} \hline 1 & 2 & \\ \hline 3 & 4 & \\ \hline & & \\ \hline \end{tabular}$

$\begin{tabular}{|c|c|c|} \hline 1 & 3 & \\ \hline 2 & 4 & \\ \hline & & \\ \hline \end{tabular}$

Second, let's look at the numbers that form a vertical "L", sometimes called tetris $J$:

$\begin{tabular}{|c|c|c|} \hline 1 & 4 & \\ \hline 2 & & \\ \hline 3 & & \\ \hline \end{tabular}$

$\begin{tabular}{|c|c|c|} \hline 1 & 3 & \\ \hline 2 & & \\ \hline 4 & & \\ \hline \end{tabular}$

$\begin{tabular}{|c|c|c|} \hline 1 & 2 & \\ \hline 3 & & \\ \hline 4 & & \\ \hline \end{tabular}$

Third, let's look at the numbers that form a horizontal "L", sometimes called tetris $L$:

$\begin{tabular}{|c|c|c|} \hline 1 & 2 & 3 \\ \hline 4 & & \\ \hline & & \\ \hline \end{tabular}$

$\begin{tabular}{|c|c|c|} \hline 1 & 2 & 4 \\ \hline 3 & & \\ \hline & & \\ \hline \end{tabular}$

$\begin{tabular}{|c|c|c|} \hline 1 & 3 & 4 \\ \hline 2 & & \\ \hline & & \\ \hline \end{tabular}$

Now, the numbers 6-9 will form similar shapes (rotated by 180 degrees, and anchored in the lower-right corner of the 3x3 grid).

If you match up one tetris shape from the numbers 1-4 and one tetris shape from the numbers 6-9, there is only one place left for the number 5 to be placed.

So what shapes will physically fit in the 3x3 grid, together?

$\begin{array}{ccl} 1 - 4 \text{ shape} & 6 - 9 \text{ shape} & \text{number of pairings} \\ O & J & 2\times 3 = 6 \\ O & L & 2\times 3 = 6 \\ J & O & 3\times 2 = 6 \\ J & J & 3 \times 3 = 9 \\ L & O & 3 \times 2 = 6 \\ L & L & 3 \times 3 = 9 \\ O & O & \qquad \text{They don't fit} \\ J & L & \qquad \text{They don't fit} \\ L & J & \qquad \text{They don't fit} \\ \end{array}$

The answer is $4\times 6 + 2\times 9 = \boxed{\text{(D) }42}$.