Number of ways of coloring n objects which are laid in a row with k colors such that the adjacent objects are...
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 21 hours ago
Brij Raj Kishore
20012
add a comment |
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 21 hours ago
Brij Raj Kishore
20012
What you want is the chromatic polynomial for the path graph.
– Gerry Myerson
21 hours ago
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
– Brij Raj Kishore
21 hours ago
add a comment |
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 21 hours ago
Brij Raj Kishore
20012
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 21 hours ago
Brij Raj Kishore
20012
asked 21 hours ago
Brij Raj Kishore
20012
asked 21 hours ago
Brij Raj Kishore
20012
asked 21 hours ago
Brij Raj Kishore
20012
asked 21 hours ago
Brij Raj Kishore
20012
20012
What you want is the chromatic polynomial for the path graph.
– Gerry Myerson
21 hours ago
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
– Brij Raj Kishore
21 hours ago
add a comment |
What you want is the chromatic polynomial for the path graph.
– Gerry Myerson
21 hours ago
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
– Brij Raj Kishore
21 hours ago
What you want is the chromatic polynomial for the path graph.
– Gerry Myerson
21 hours ago
What you want is the chromatic polynomial for the path graph.
– Gerry Myerson
21 hours ago
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
– Brij Raj Kishore
21 hours ago
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
– Brij Raj Kishore
21 hours ago
add a comment |
1 Answer
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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 21 hours ago
Oldboy
7,1191832
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 21 hours ago
Oldboy
7,1191832
add a comment |
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 21 hours ago
Oldboy
7,1191832
add a comment |
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 21 hours ago
Oldboy
7,1191832
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 21 hours ago
Oldboy
7,1191832
answered 21 hours ago
Oldboy
7,1191832
answered 21 hours ago
Oldboy
7,1191832
answered 21 hours ago
Oldboy
7,1191832
7,1191832
add a comment |
add a comment |
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What you want is the chromatic polynomial for the path graph.
– Gerry Myerson
21 hours ago
I guess so. This is the chromatic polynomial as the adjacent colors have to be different
– Brij Raj Kishore
21 hours ago