Mixing
How many ways to color each cell of $2times 13$ table with $4$ colors so that cells with a common side aren't...
$begingroup$
$2times 13$ table cells can be painted red, blue, green and yellow. "Neighbor" cells with common sides must have different colors.
How many ways to do this are there?
Thanks in a lot advance!
combinatorics
asked Jan 20 at 14:24
Xxx DddXxx Ddd
354
$endgroup$
add a comment |
$begingroup$
$2times 13$ table cells can be painted red, blue, green and yellow. "Neighbor" cells with common sides must have different colors.
How many ways to do this are there?
Thanks in a lot advance!
combinatorics
asked Jan 20 at 14:24
Xxx DddXxx Ddd
354
$endgroup$
add a comment |
$begingroup$
$2times 13$ table cells can be painted red, blue, green and yellow. "Neighbor" cells with common sides must have different colors.
How many ways to do this are there?
Thanks in a lot advance!
combinatorics
asked Jan 20 at 14:24
Xxx DddXxx Ddd
354
$endgroup$
$2times 13$ table cells can be painted red, blue, green and yellow. "Neighbor" cells with common sides must have different colors.
How many ways to do this are there?
Thanks in a lot advance!
combinatorics
combinatorics
asked Jan 20 at 14:24
Xxx DddXxx Ddd
354
asked Jan 20 at 14:24
Xxx DddXxx Ddd
354
asked Jan 20 at 14:24
Xxx DddXxx Ddd
354
asked Jan 20 at 14:24
Xxx DddXxx Ddd
354
asked Jan 20 at 14:24
Xxx DddXxx Ddd
354
354
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.
answered Jan 20 at 14:50
Christian BlatterChristian Blatter
174k8115327
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$begingroup$
There are $4times3=12$ possibilities for the first column.
If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.
This leads to $12times7^{12}$ possibilities.
answered Jan 20 at 14:51
drhabdrhab
102k545136
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add a comment |
$begingroup$
A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.
If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.
Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).
Then we proceed as follows:
- First Column in 12 ways:
- Choose 1 in 4 ways.
- Choose 14 in 3 ways.
- Adjacent Columns in $3^{12} times 5^{12}$ ways:
- Choose 2 to 13 in $3^{12}$ ways.
- Choose 15 to 26 in $5^{12}$ ways:
- 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,
- 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.
So we can choose in $12 times 3^{12} times {5}^{12}$ ways.
answered Jan 20 at 15:08
SagnikSagnik
1377
$endgroup$
1
$begingroup$
You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
$endgroup$
– Ross Millikan
Jan 20 at 15:12
$begingroup$
I stated 4 colors in the post, thanks anyway
$endgroup$
– Xxx Ddd
Jan 20 at 15:13
$begingroup$
@Ross right... I hadn't considered that.
$endgroup$
– Sagnik
Jan 20 at 15:18
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
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$begingroup$
Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.
answered Jan 20 at 14:50
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.
answered Jan 20 at 14:50
Christian BlatterChristian Blatter
174k8115327
$endgroup$
add a comment |
$begingroup$
Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.
answered Jan 20 at 14:50
Christian BlatterChristian Blatter
174k8115327
$endgroup$
Find out in how many ways you can color the two leftmost fields $(1,1)/(2,1)$. Then assume that all fields $(1,j)/(2,j)$ $>(1leq jleq r)$ have been admissibly colored, and find out in how many ways you can color the fields $(1,r+1)/(2,r+1)$. This will lead you to a simple recursion.
answered Jan 20 at 14:50
Christian BlatterChristian Blatter
174k8115327
answered Jan 20 at 14:50
Christian BlatterChristian Blatter
174k8115327
answered Jan 20 at 14:50
Christian BlatterChristian Blatter
174k8115327
answered Jan 20 at 14:50
Christian BlatterChristian Blatter
174k8115327
174k8115327
add a comment |
add a comment |
$begingroup$
There are $4times3=12$ possibilities for the first column.
If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.
This leads to $12times7^{12}$ possibilities.
answered Jan 20 at 14:51
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
There are $4times3=12$ possibilities for the first column.
If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.
This leads to $12times7^{12}$ possibilities.
answered Jan 20 at 14:51
drhabdrhab
102k545136
$endgroup$
add a comment |
$begingroup$
There are $4times3=12$ possibilities for the first column.
If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.
This leads to $12times7^{12}$ possibilities.
answered Jan 20 at 14:51
drhabdrhab
102k545136
$endgroup$
There are $4times3=12$ possibilities for the first column.
If $nin{1,dots,12}$ consecutive columns among which the first column have been painted then there are $1times3+2times2=7$ possibilities to paint the $n+1$-th bordering column.
This leads to $12times7^{12}$ possibilities.
answered Jan 20 at 14:51
drhabdrhab
102k545136
answered Jan 20 at 14:51
drhabdrhab
102k545136
answered Jan 20 at 14:51
drhabdrhab
102k545136
answered Jan 20 at 14:51
drhabdrhab
102k545136
102k545136
add a comment |
add a comment |
$begingroup$
A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.
If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.
Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).
Then we proceed as follows:
- First Column in 12 ways:
- Choose 1 in 4 ways.
- Choose 14 in 3 ways.
- Adjacent Columns in $3^{12} times 5^{12}$ ways:
- Choose 2 to 13 in $3^{12}$ ways.
- Choose 15 to 26 in $5^{12}$ ways:
- 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,
- 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.
So we can choose in $12 times 3^{12} times {5}^{12}$ ways.
answered Jan 20 at 15:08
SagnikSagnik
1377
$endgroup$
1
$begingroup$
You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
$endgroup$
– Ross Millikan
Jan 20 at 15:12
$begingroup$
I stated 4 colors in the post, thanks anyway
$endgroup$
– Xxx Ddd
Jan 20 at 15:13
$begingroup$
@Ross right... I hadn't considered that.
$endgroup$
– Sagnik
Jan 20 at 15:18
add a comment |
$begingroup$
A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.
If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.
Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).
Then we proceed as follows:
- First Column in 12 ways:
- Choose 1 in 4 ways.
- Choose 14 in 3 ways.
- Adjacent Columns in $3^{12} times 5^{12}$ ways:
- Choose 2 to 13 in $3^{12}$ ways.
- Choose 15 to 26 in $5^{12}$ ways:
- 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,
- 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.
So we can choose in $12 times 3^{12} times {5}^{12}$ ways.
answered Jan 20 at 15:08
SagnikSagnik
1377
$endgroup$
1
$begingroup$
You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
$endgroup$
– Ross Millikan
Jan 20 at 15:12
$begingroup$
I stated 4 colors in the post, thanks anyway
$endgroup$
– Xxx Ddd
Jan 20 at 15:13
$begingroup$
@Ross right... I hadn't considered that.
$endgroup$
– Sagnik
Jan 20 at 15:18
add a comment |
$begingroup$
A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.
If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.
Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).
Then we proceed as follows:
- First Column in 12 ways:
- Choose 1 in 4 ways.
- Choose 14 in 3 ways.
- Adjacent Columns in $3^{12} times 5^{12}$ ways:
- Choose 2 to 13 in $3^{12}$ ways.
- Choose 15 to 26 in $5^{12}$ ways:
- 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,
- 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.
So we can choose in $12 times 3^{12} times {5}^{12}$ ways.
answered Jan 20 at 15:08
SagnikSagnik
1377
$endgroup$
A table with 2 rows and 13 columns can be interpreted as a graph where each cell is a vertex having edges to the cell beside it and below it.
If $(1,1)$ cell is marked as $textbf{1}$ , $(1,2)$ cell is marked as $textbf{2}$ and so on then the $(2,13)$ cell is marked as 26.
Note that Vertex 1 has 4 choices. Vertices 2 and 14 have 3 choices each. Now vertex 15 has 2 choices (if 2 and 14 have the same color) or 3 choices (if 2 and 14 have different colors).
Then we proceed as follows:
- First Column in 12 ways:
- Choose 1 in 4 ways.
- Choose 14 in 3 ways.
- Adjacent Columns in $3^{12} times 5^{12}$ ways:
- Choose 2 to 13 in $3^{12}$ ways.
- Choose 15 to 26 in $5^{12}$ ways:
- 2 ways. Consider 2 and 14 have different colors, then we have $binom{2}{1}$ colors to choose from, or,
- 3 ways. Consider 2 and 14 have the same colors, then we have $binom{3}{1}$ colors to choose from.
So we can choose in $12 times 3^{12} times {5}^{12}$ ways.
answered Jan 20 at 15:08
SagnikSagnik
1377
edited Jan 20 at 15:55
answered Jan 20 at 15:08
SagnikSagnik
1377
answered Jan 20 at 15:08
SagnikSagnik
1377
answered Jan 20 at 15:08
SagnikSagnik
1377
1377
1
$begingroup$
You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
$endgroup$
– Ross Millikan
Jan 20 at 15:12
$begingroup$
I stated 4 colors in the post, thanks anyway
$endgroup$
– Xxx Ddd
Jan 20 at 15:13
$begingroup$
@Ross right... I hadn't considered that.
$endgroup$
– Sagnik
Jan 20 at 15:18
add a comment |
1
$begingroup$
You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
$endgroup$
– Ross Millikan
Jan 20 at 15:12
$begingroup$
I stated 4 colors in the post, thanks anyway
$endgroup$
– Xxx Ddd
Jan 20 at 15:13
$begingroup$
@Ross right... I hadn't considered that.
$endgroup$
– Sagnik
Jan 20 at 15:18
1
1
$begingroup$
You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
$endgroup$
– Ross Millikan
Jan 20 at 15:12
$begingroup$
You seem to be assuming that $1,3,ldots 15$ all have to have the same color. That was not specified.
$endgroup$
– Ross Millikan
Jan 20 at 15:12
$begingroup$
I stated 4 colors in the post, thanks anyway
$endgroup$
– Xxx Ddd
Jan 20 at 15:13
$begingroup$
I stated 4 colors in the post, thanks anyway
$endgroup$
– Xxx Ddd
Jan 20 at 15:13
$begingroup$
@Ross right... I hadn't considered that.
$endgroup$
– Sagnik
Jan 20 at 15:18
$begingroup$
@Ross right... I hadn't considered that.
$endgroup$
– Sagnik
Jan 20 at 15:18
add a comment |
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