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

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{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #17]] and [[2010 AMC 10B Problems|2010 AMC 10B #23]]}}
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== Problem ==
 
== Problem ==
 
The entries in a <math>3 \times 3</math> array include all the digits from <math>1</math> through <math>9</math>, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
 
The entries in a <math>3 \times 3</math> array include all the digits from <math>1</math> through <math>9</math>, arranged so that the entries in every row and column are in increasing order. How many such arrays are there?
Line 4: Line 6:
 
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60</math>
 
<math>\textbf{(A)}\ 18 \qquad \textbf{(B)}\ 24 \qquad \textbf{(C)}\ 36 \qquad \textbf{(D)}\ 42 \qquad \textbf{(E)}\ 60</math>
  
== Solution ==
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== Solution 1 ==
The first 4 numbers will form one of 3 tetris "shapes".
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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.  
  
First, let's look at the numbers that form a 2x2 block, sometimes called tetris <math> O</math>:
 
  
<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & \\
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*'''Case 1: Center 4'''
\hline 3 & 4 & \\
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<cmath>\begin{tabular}{|c|c|c|} \hline 1&2&\\
\hline & & \\
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\hline 3&4&8\\
\hline \end{tabular}</math>
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\hline &&9\\
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\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\
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\hline 3&4&\\
 +
\hline &8&9\\
 +
\hline \end{tabular}</cmath>
  
<math> \begin{tabular}{|c|c|c|} \hline 1 & 3 & \\
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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. <math>2*6=12</math>
\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>:
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*'''Case 2: Center 5'''
 +
<cmath>\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\\
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\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&3\\
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\hline 4&5&8\\
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\hline &&9\\
 +
\hline \end{tabular}</cmath>
  
<math> \begin{tabular}{|c|c|c|} \hline 1 & 4 & \\
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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 <math>4<5</math>, logically see that the numbers of cases are then 2,3,3,1 respectively. By symmetry, <math>2*9=18</math>
\hline 2 & & \\
 
\hline 3 & & \\
 
\hline \end{tabular}</math>
 
  
<math> \begin{tabular}{|c|c|c|} \hline 1 & 3 & \\
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*'''Case 3: Center 6'''
\hline 2 & & \\
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By inspection, realize that this is symmetrical to case 1 except that the 7s instead of the 3s are assured.
\hline 4 & & \\
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<math>2*6=12</math>
\hline \end{tabular}</math>
 
  
<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & \\
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<cmath>12+18+12=\boxed{\textbf{D)}42}</cmath>
\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 \\
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~BJHHar
\hline 4 & & \\
 
\hline & & \\
 
\hline \end{tabular}</math>
 
  
<math> \begin{tabular}{|c|c|c|} \hline 1 & 2 & 4 \\
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== Solution 2==
\hline 3 & & \\
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This solution is trivial by the hook length theorem. The hooks look like this:
\hline & & \\
 
\hline \end{tabular}</math>
 
  
<math> \begin{tabular}{|c|c|c|} \hline 1 & 3 & 4 \\
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<math> \begin{tabular}{|c|c|c|} \hline 5 & 4 & 3 \\
\hline 2 & & \\
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\hline 4 & 3 & 2\\
\hline & & \\
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\hline 3 & 2 & 1\\
 
\hline \end{tabular}</math>
 
\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).
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So, the answer is <math>\frac{9!}{5 \cdot 4 \cdot 3 \cdot 4 \cdot 3 \cdot 2 \cdot 3 \cdot 2 \cdot 1}</math> = <math>\boxed{\text{(D) }42}</math>
 
 
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?
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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.
  
<math> \begin{array}{ccl} 1 - 4 shape & 6 - 9 shape & number of pairings \\
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==Video Solution==
O & J & 2\times 3 = 6 \\
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https://youtu.be/ZfnxbpdFKjU?t=422
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>.
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~IceMatrix
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=16|num-a=18|ab=B}}
 
{{AMC12 box|year=2010|num-b=16|num-a=18|ab=B}}
 +
{{AMC10 box|year=2010|num-b=22|num-a=24|ab=B}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 19:41, 22 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

Solution 2

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

See also

2010 AMC 12B (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2010 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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