Number of ways of coloring n objects which are laid in a row with k colors such that the adjacent objects are...
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Given n objects, which are lying in a straight line next to each other, in how many ways we can color them with k colors (all must be painted) such that the adjacent boxes not of same colors.
I can feel that inclusion exclusion principle will apply here but I am not able to figure out where to start. Its been a while since I read about them.
combinatorics inclusion-exclusion coloring
asked Jan 4 at 15:55
Brij Raj KishoreBrij Raj Kishore
215112
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add a comment |
$begingroup$
Given n objects, which are lying in a straight line next to each other, in how many ways we can color them with k colors (all must be painted) such that the adjacent boxes not of same colors.
I can feel that inclusion exclusion principle will apply here but I am not able to figure out where to start. Its been a while since I read about them.
combinatorics inclusion-exclusion coloring
asked Jan 4 at 15:55
Brij Raj KishoreBrij Raj Kishore
215112
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$begingroup$
What you want is the chromatic polynomial for the path graph.
$endgroup$
– Gerry Myerson
Jan 4 at 15:58
$begingroup$
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
$endgroup$
– Brij Raj Kishore
Jan 4 at 16:03
add a comment |
$begingroup$
Given n objects, which are lying in a straight line next to each other, in how many ways we can color them with k colors (all must be painted) such that the adjacent boxes not of same colors.
I can feel that inclusion exclusion principle will apply here but I am not able to figure out where to start. Its been a while since I read about them.
combinatorics inclusion-exclusion coloring
asked Jan 4 at 15:55
Brij Raj KishoreBrij Raj Kishore
215112
$endgroup$
Given n objects, which are lying in a straight line next to each other, in how many ways we can color them with k colors (all must be painted) such that the adjacent boxes not of same colors.
I can feel that inclusion exclusion principle will apply here but I am not able to figure out where to start. Its been a while since I read about them.
combinatorics inclusion-exclusion coloring
combinatorics inclusion-exclusion coloring
asked Jan 4 at 15:55
Brij Raj KishoreBrij Raj Kishore
215112
asked Jan 4 at 15:55
Brij Raj KishoreBrij Raj Kishore
215112
asked Jan 4 at 15:55
Brij Raj KishoreBrij Raj Kishore
215112
asked Jan 4 at 15:55
Brij Raj KishoreBrij Raj Kishore
215112
asked Jan 4 at 15:55
Brij Raj KishoreBrij Raj Kishore
215112
215112
$begingroup$
What you want is the chromatic polynomial for the path graph.
$endgroup$
– Gerry Myerson
Jan 4 at 15:58
$begingroup$
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
$endgroup$
– Brij Raj Kishore
Jan 4 at 16:03
add a comment |
$begingroup$
What you want is the chromatic polynomial for the path graph.
$endgroup$
– Gerry Myerson
Jan 4 at 15:58
$begingroup$
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
$endgroup$
– Brij Raj Kishore
Jan 4 at 16:03
$begingroup$
What you want is the chromatic polynomial for the path graph.
$endgroup$
– Gerry Myerson
Jan 4 at 15:58
$begingroup$
What you want is the chromatic polynomial for the path graph.
$endgroup$
– Gerry Myerson
Jan 4 at 15:58
$begingroup$
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
$endgroup$
– Brij Raj Kishore
Jan 4 at 16:03
$begingroup$
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
$endgroup$
– Brij Raj Kishore
Jan 4 at 16:03
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
You can color the first box in any of $k$ colors availble to you. The second box can be colored with one of the remaining $k-1$ colors. The same is true for the third, fourth... So the total number of colorings is $ktimes(k-1)^{n-1}$
answered Jan 4 at 16:04
OldboyOldboy
9,26611138
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add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
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oldest
votes
active
oldest
votes
$begingroup$
You can color the first box in any of $k$ colors availble to you. The second box can be colored with one of the remaining $k-1$ colors. The same is true for the third, fourth... So the total number of colorings is $ktimes(k-1)^{n-1}$
answered Jan 4 at 16:04
OldboyOldboy
9,26611138
$endgroup$
add a comment |
$begingroup$
You can color the first box in any of $k$ colors availble to you. The second box can be colored with one of the remaining $k-1$ colors. The same is true for the third, fourth... So the total number of colorings is $ktimes(k-1)^{n-1}$
answered Jan 4 at 16:04
OldboyOldboy
9,26611138
$endgroup$
add a comment |
$begingroup$
You can color the first box in any of $k$ colors availble to you. The second box can be colored with one of the remaining $k-1$ colors. The same is true for the third, fourth... So the total number of colorings is $ktimes(k-1)^{n-1}$
answered Jan 4 at 16:04
OldboyOldboy
9,26611138
$endgroup$
You can color the first box in any of $k$ colors availble to you. The second box can be colored with one of the remaining $k-1$ colors. The same is true for the third, fourth... So the total number of colorings is $ktimes(k-1)^{n-1}$
answered Jan 4 at 16:04
OldboyOldboy
9,26611138
answered Jan 4 at 16:04
OldboyOldboy
9,26611138
answered Jan 4 at 16:04
OldboyOldboy
9,26611138
answered Jan 4 at 16:04
OldboyOldboy
9,26611138
9,26611138
add a comment |
add a comment |
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$begingroup$
What you want is the chromatic polynomial for the path graph.
$endgroup$
– Gerry Myerson
Jan 4 at 15:58
$begingroup$
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
$endgroup$
– Brij Raj Kishore
Jan 4 at 16:03