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

Revision as of 12:53, 13 December 2020

The following problem is from both the 2010 AMC 12B #17 and 2010 AMC 10B #23, so both problems redirect to this page.

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 1

Observe that all tableaus must have 1s and 9s in the corners, 8s and 2s next to those corner squares, and 4-6 in the middle square. Also note that for each tableau, there exists a valid tableau diagonally symmetrical across the diagonal extending from the top left to the bottom right.

Case 1: Center 4

$\begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&4&\\ \hline &8&9\\ \hline \end{tabular}$

3 necessarily must be placed as above. Any number could fill the isolated square, but the other 2 are then invariant. So, there are 3 cases each and 6 overall cases. Given diagonal symmetry, alternate 2 and 8 placements yield symmetrical cases. $2*6=12$

Case 2: Center 5

$\begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&\\ \hline &8&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\ \hline 3&5&8\\ \hline &&9\\ \hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\ \hline 4&5&8\\ \hline &&9\\ \hline \end{tabular}$

Here, no 3s or 7s are assured, but this is only a teensy bit trickier and messier. WLOG, casework with 3 instead of 7 as above. Remembering that $4<5$ , logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, $2*9=18$

Case 3: Center 6

By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured. $2*6=12$

$12+18+12=\boxed{\textbf{D)}42}$

~BJHHar

P.S.: I like the tetris approach used in Solution 2 but found it a bit arbitrary. Solution 3 is the best, but not many would know hook length theorem. If the initial observations are unclear, make a tableau with a range of possible numbers in each square.

Solution 3

This solution is trivial by the hook length theorem. The hooks look like this:

$\begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\ \hline 4 & 3 & 2\\ \hline 3 & 2 & 1\\ \hline \end{tabular}$

So, the answer is $\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}$ = $\boxed{\text{(D) }42}$

P.S. The hook length formula is a formula to calculate the number of standard Young tableaux of a Young diagram. Numberphile has an easy-to-understand video about it here: https://www.youtube.com/watch?v=vgZhrEs4tuk The full proof is quite complicated and is not given in the video, although the video hints at possible proofs.

Video Solution

https://youtu.be/ZfnxbpdFKjU?t=422

~IceMatrix

@@ Line 49: / Line 49: @@
 ''P.S.: I like the tetris approach used in Solution 2 but found it a bit arbitrary. Solution 3 is the best, but not many would know hook length theorem. If the initial observations are unclear, make a tableau with a range of possible numbers in each square''.
-== Solution 2==
-The first 4 numbers will form one of 3 tetris "shapes".
-First, let's look at the numbers that form a <math>2\times2</math> block, sometimes called tetris <math> O</math>:
-<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & \\
-\hline 3 & 4 & \\
-\hline & & \\
-\hline \end{tabular}</math>
-<math> \begin{tabular}{|c|c|c|} \hline 1 & 3 & \\
-\hline 2 & 4 & \\
-\hline & & \\
-\hline \end{tabular}</math>
-Second, let's look at the numbers that form a vertical "L", sometimes called tetris <math> J</math>:
-<math> \begin{tabular}{|c|c|c|} \hline 1 & 4 & \\
-\hline 2 & & \\
-\hline 3 & & \\
-\hline \end{tabular}</math>
-<math> \begin{tabular}{|c|c|c|} \hline 1 & 3 & \\
-\hline 2 & & \\
-\hline 4 & & \\
-\hline \end{tabular}</math>
-<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & \\
-\hline 3 & & \\
-\hline 4 & & \\
-\hline \end{tabular}</math>
-Third, let's look at the numbers that form a horizontal "L", sometimes called tetris <math> L</math>:
-<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & 3 \\
-\hline 4 & & \\
-\hline & & \\
-\hline \end{tabular}</math>
-<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & 4 \\
-\hline 3 & & \\
-\hline & & \\
-\hline \end{tabular}</math>
-<math> \begin{tabular}{|c|c|c|} \hline 1 & 3 & 4 \\
-\hline 2 & & \\
-\hline & & \\
-\hline \end{tabular}</math>
-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?
-<math> \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}</math>
-The answer is <math> 4\times 6 + 2\times 9 = \boxed{\text{(D) }42}</math>.
 == Solution 3==

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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