Derrangments function for coloring a chess table
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How many ways are there to color a chess table of size n*n with n different colors. We color in such a way that in each horizontal row there are all colors and at the same time in no vertical row there are two fields of the same color next to each other. I think we can use derangements for this but I'm not sure how.
combinatorics discrete-mathematics
asked Dec 10 '18 at 15:43
ponikoliponikoli
416
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add a comment |
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How many ways are there to color a chess table of size n*n with n different colors. We color in such a way that in each horizontal row there are all colors and at the same time in no vertical row there are two fields of the same color next to each other. I think we can use derangements for this but I'm not sure how.
combinatorics discrete-mathematics
asked Dec 10 '18 at 15:43
ponikoliponikoli
416
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Where did this question come from?
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– coffeemath
Dec 10 '18 at 15:46
add a comment |
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How many ways are there to color a chess table of size n*n with n different colors. We color in such a way that in each horizontal row there are all colors and at the same time in no vertical row there are two fields of the same color next to each other. I think we can use derangements for this but I'm not sure how.
combinatorics discrete-mathematics
asked Dec 10 '18 at 15:43
ponikoliponikoli
416
$endgroup$
How many ways are there to color a chess table of size n*n with n different colors. We color in such a way that in each horizontal row there are all colors and at the same time in no vertical row there are two fields of the same color next to each other. I think we can use derangements for this but I'm not sure how.
combinatorics discrete-mathematics
combinatorics discrete-mathematics
asked Dec 10 '18 at 15:43
ponikoliponikoli
416
asked Dec 10 '18 at 15:43
ponikoliponikoli
416
asked Dec 10 '18 at 15:43
ponikoliponikoli
416
asked Dec 10 '18 at 15:43
ponikoliponikoli
416
asked Dec 10 '18 at 15:43
ponikoliponikoli
416
416
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Where did this question come from?
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– coffeemath
Dec 10 '18 at 15:46
add a comment |
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Where did this question come from?
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– coffeemath
Dec 10 '18 at 15:46
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Where did this question come from?
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– coffeemath
Dec 10 '18 at 15:46
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Where did this question come from?
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– coffeemath
Dec 10 '18 at 15:46
add a comment |
1 Answer
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Fill in the first row any way you like. How many options? Now, you are correct, the second row has to be a derangement of the first. How many options there? Each row in turn is a derangement of the one above.
answered Dec 10 '18 at 15:48
Ross MillikanRoss Millikan
297k23198371
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So for the first row I think there are n^n ways we can color the first row. So for the second row there are D(n^n) ways to color it? But how do I continue this?
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– ponikoli
Dec 10 '18 at 19:20
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No, you need one of each color in the first row, so there are fewer. How many?
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– Ross Millikan
Dec 10 '18 at 19:35
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There is n^n * n^(n-1) * n^(n-2) * n^(n-3) ... n^1.
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– ponikoli
Dec 10 '18 at 19:44
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The first row has $n$ objects that you can put in any order, so there are $n!$ ways to do it.
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– Ross Millikan
Dec 10 '18 at 20:52
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Then the second row has D(n!) ways to do it. The third D(D(n!)) etc..?
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– ponikoli
Dec 11 '18 at 9:00
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Your Answer
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1 Answer
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1 Answer
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oldest
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votes
$begingroup$
Fill in the first row any way you like. How many options? Now, you are correct, the second row has to be a derangement of the first. How many options there? Each row in turn is a derangement of the one above.
answered Dec 10 '18 at 15:48
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
So for the first row I think there are n^n ways we can color the first row. So for the second row there are D(n^n) ways to color it? But how do I continue this?
$endgroup$
– ponikoli
Dec 10 '18 at 19:20
$begingroup$
No, you need one of each color in the first row, so there are fewer. How many?
$endgroup$
– Ross Millikan
Dec 10 '18 at 19:35
$begingroup$
There is n^n * n^(n-1) * n^(n-2) * n^(n-3) ... n^1.
$endgroup$
– ponikoli
Dec 10 '18 at 19:44
$begingroup$
The first row has $n$ objects that you can put in any order, so there are $n!$ ways to do it.
$endgroup$
– Ross Millikan
Dec 10 '18 at 20:52
$begingroup$
Then the second row has D(n!) ways to do it. The third D(D(n!)) etc..?
$endgroup$
– ponikoli
Dec 11 '18 at 9:00
|
show 6 more comments
$begingroup$
Fill in the first row any way you like. How many options? Now, you are correct, the second row has to be a derangement of the first. How many options there? Each row in turn is a derangement of the one above.
answered Dec 10 '18 at 15:48
Ross MillikanRoss Millikan
297k23198371
$endgroup$
$begingroup$
So for the first row I think there are n^n ways we can color the first row. So for the second row there are D(n^n) ways to color it? But how do I continue this?
$endgroup$
– ponikoli
Dec 10 '18 at 19:20
$begingroup$
No, you need one of each color in the first row, so there are fewer. How many?
$endgroup$
– Ross Millikan
Dec 10 '18 at 19:35
$begingroup$
There is n^n * n^(n-1) * n^(n-2) * n^(n-3) ... n^1.
$endgroup$
– ponikoli
Dec 10 '18 at 19:44
$begingroup$
The first row has $n$ objects that you can put in any order, so there are $n!$ ways to do it.
$endgroup$
– Ross Millikan
Dec 10 '18 at 20:52
$begingroup$
Then the second row has D(n!) ways to do it. The third D(D(n!)) etc..?
$endgroup$
– ponikoli
Dec 11 '18 at 9:00
|
show 6 more comments
$begingroup$
Fill in the first row any way you like. How many options? Now, you are correct, the second row has to be a derangement of the first. How many options there? Each row in turn is a derangement of the one above.
answered Dec 10 '18 at 15:48
Ross MillikanRoss Millikan
297k23198371
$endgroup$
Fill in the first row any way you like. How many options? Now, you are correct, the second row has to be a derangement of the first. How many options there? Each row in turn is a derangement of the one above.
answered Dec 10 '18 at 15:48
Ross MillikanRoss Millikan
297k23198371
answered Dec 10 '18 at 15:48
Ross MillikanRoss Millikan
297k23198371
answered Dec 10 '18 at 15:48
Ross MillikanRoss Millikan
297k23198371
answered Dec 10 '18 at 15:48
Ross MillikanRoss Millikan
297k23198371
297k23198371
$begingroup$
So for the first row I think there are n^n ways we can color the first row. So for the second row there are D(n^n) ways to color it? But how do I continue this?
$endgroup$
– ponikoli
Dec 10 '18 at 19:20
$begingroup$
No, you need one of each color in the first row, so there are fewer. How many?
$endgroup$
– Ross Millikan
Dec 10 '18 at 19:35
$begingroup$
There is n^n * n^(n-1) * n^(n-2) * n^(n-3) ... n^1.
$endgroup$
– ponikoli
Dec 10 '18 at 19:44
$begingroup$
The first row has $n$ objects that you can put in any order, so there are $n!$ ways to do it.
$endgroup$
– Ross Millikan
Dec 10 '18 at 20:52
$begingroup$
Then the second row has D(n!) ways to do it. The third D(D(n!)) etc..?
$endgroup$
– ponikoli
Dec 11 '18 at 9:00
|
show 6 more comments
$begingroup$
So for the first row I think there are n^n ways we can color the first row. So for the second row there are D(n^n) ways to color it? But how do I continue this?
$endgroup$
– ponikoli
Dec 10 '18 at 19:20
$begingroup$
No, you need one of each color in the first row, so there are fewer. How many?
$endgroup$
– Ross Millikan
Dec 10 '18 at 19:35
$begingroup$
There is n^n * n^(n-1) * n^(n-2) * n^(n-3) ... n^1.
$endgroup$
– ponikoli
Dec 10 '18 at 19:44
$begingroup$
The first row has $n$ objects that you can put in any order, so there are $n!$ ways to do it.
$endgroup$
– Ross Millikan
Dec 10 '18 at 20:52
$begingroup$
Then the second row has D(n!) ways to do it. The third D(D(n!)) etc..?
$endgroup$
– ponikoli
Dec 11 '18 at 9:00
$begingroup$
So for the first row I think there are n^n ways we can color the first row. So for the second row there are D(n^n) ways to color it? But how do I continue this?
$endgroup$
– ponikoli
Dec 10 '18 at 19:20
$begingroup$
So for the first row I think there are n^n ways we can color the first row. So for the second row there are D(n^n) ways to color it? But how do I continue this?
$endgroup$
– ponikoli
Dec 10 '18 at 19:20
$begingroup$
No, you need one of each color in the first row, so there are fewer. How many?
$endgroup$
– Ross Millikan
Dec 10 '18 at 19:35
$begingroup$
No, you need one of each color in the first row, so there are fewer. How many?
$endgroup$
– Ross Millikan
Dec 10 '18 at 19:35
$begingroup$
There is n^n * n^(n-1) * n^(n-2) * n^(n-3) ... n^1.
$endgroup$
– ponikoli
Dec 10 '18 at 19:44
$begingroup$
There is n^n * n^(n-1) * n^(n-2) * n^(n-3) ... n^1.
$endgroup$
– ponikoli
Dec 10 '18 at 19:44
$begingroup$
The first row has $n$ objects that you can put in any order, so there are $n!$ ways to do it.
$endgroup$
– Ross Millikan
Dec 10 '18 at 20:52
$begingroup$
The first row has $n$ objects that you can put in any order, so there are $n!$ ways to do it.
$endgroup$
– Ross Millikan
Dec 10 '18 at 20:52
$begingroup$
Then the second row has D(n!) ways to do it. The third D(D(n!)) etc..?
$endgroup$
– ponikoli
Dec 11 '18 at 9:00
$begingroup$
Then the second row has D(n!) ways to do it. The third D(D(n!)) etc..?
$endgroup$
– ponikoli
Dec 11 '18 at 9:00
|
show 6 more comments
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– coffeemath
Dec 10 '18 at 15:46