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 5: 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&\\
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\hline &8&9\\
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\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'''
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<cmath>\begin{tabular}{|c|c|c|} \hline 1&2&3\\
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\hline 4&5&\\
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\hline &8&9\\
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\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\
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\hline 3&5&\\
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\hline &8&9\\
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\hline \end{tabular} \;\;\; \begin{tabular}{|c|c|c|} \hline 1&2&\\
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\hline 3&5&8\\
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\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\\
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\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.
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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.
  
So what shapes will physically fit in the 3x3 grid, together?
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==Video Solution==
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https://youtu.be/ZfnxbpdFKjU?t=422
  
<math> \begin{tabular}{ccl} 1 - 4 shape & 6 - 9 shape & number of pairings \\
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~IceMatrix
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{tabular}</math>
 
 
 
The answer is <math> 4\times 6 + 2\times 9 = \boxed{\text{(D) }42}</math>.
 
  
 
== 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]]
 +
{{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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