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Difference between revisions of "2008 UNCO Math Contest II Problems/Problem 1"

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== Solution ==  
 
== Solution ==  
 
{{solution}}
 
{{solution}}
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Case 1: The numbers include <math>0,1,2</math>
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Row 1: <math>0 1 1</math>
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Row 2: <math>2 0 0</math>
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Row 3: <math>0 1 1</math>
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There are 3 rows to choose where to place Row 2, giving us <math>\boxed3</math> cases.
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Case 2: The numbers include <math>0,1,2</math>
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Row 1: <math>1 0 1</math>
 +
 +
Row 2: <math>0 2 0</math>
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 +
Row 3: <math>1 0 1</math>
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There are 3 rows to choose where to place Row 2, giving us <math>\boxed3</math> cases.
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Case 3: The numbers include <math>0, 1, 2</math>
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Row 1: <math>1 1 0</math>
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 +
Row 2: <math>0 0 2</math>
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 +
Row 3: <math>1 1 0</math>
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There are 3 rows to choose where to place Row 2, giving us <math>\boxed3</math> cases.
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Case 4: The numbers include <math>0, 2</math>
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Row 1: <math>0 0 2</math>
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 +
Row 2: <math>0 2 0</math>
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Row 3: <math>2 0 0</math>
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 +
There are 3 rows to choose where to place Row 2, and Row 1 can go on either top or bottom, giving us <math>3\times2=\boxed6</math> cases.
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Case 5: The numbers include <math>0, 1</math>
 +
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Row 1: <math>0 1 1</math>
 +
 +
Row 2: <math>1 0 1</math>
 +
 +
Row 3: <math>1 1 0</math>
 +
 +
There are 3 rows to choose where to place Row 2, and Row 1 can go on either top or bottom, giving us <math>3\times2=\boxed6</math> cases.
 +
 +
This gives us a total of <math>3+3+3+6+6=\boxed{21}</math> ways.
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== See Also ==
 
== See Also ==

Revision as of 00:06, 5 December 2016

Problem

Determine the number of $3 \times 3$ square arrays whose row and column sums are equal to $2$, using $0, 1, 2$ as entries. Entries may be repeated, and not all of $0, 1, 2$ need be used as the two examples show.

\[\begin{tabular}{c c c c c c c c c} 1 & 1 & 0 & & & & 0 & 1 & 1 \\ 0 & 0 & 2 & & & & 1 & 1 & 0 \\ 1 & 1 & 0 & & & & 1 & 0 & 1 \\ \end{tabular}\]


Solution

This problem needs a solution. If you have a solution for it, please help us out by adding it.

Case 1: The numbers include $0,1,2$

Row 1: $0 1 1$

Row 2: $2 0 0$

Row 3: $0 1 1$

There are 3 rows to choose where to place Row 2, giving us $\boxed3$ cases.

Case 2: The numbers include $0,1,2$

Row 1: $1 0 1$

Row 2: $0 2 0$

Row 3: $1 0 1$

There are 3 rows to choose where to place Row 2, giving us $\boxed3$ cases.

Case 3: The numbers include $0, 1, 2$

Row 1: $1 1 0$

Row 2: $0 0 2$

Row 3: $1 1 0$

There are 3 rows to choose where to place Row 2, giving us $\boxed3$ cases.

Case 4: The numbers include $0, 2$

Row 1: $0 0 2$

Row 2: $0 2 0$

Row 3: $2 0 0$

There are 3 rows to choose where to place Row 2, and Row 1 can go on either top or bottom, giving us $3\times2=\boxed6$ cases.

Case 5: The numbers include $0, 1$

Row 1: $0 1 1$

Row 2: $1 0 1$

Row 3: $1 1 0$

There are 3 rows to choose where to place Row 2, and Row 1 can go on either top or bottom, giving us $3\times2=\boxed6$ cases.

This gives us a total of $3+3+3+6+6=\boxed{21}$ ways.



See Also

2008 UNCO Math Contest II (Problems • Answer Key • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10
All UNCO Math Contest Problems and Solutions
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