Chess Board (5×5) problemcalculating the number of possible arrangements in chesscombinatorics board with digits neat problemFinding the number of counters having each colour.Board with black and white tilesNumber of colourings for which the vertices of a square can be coloured by three different colours such that adjacent vertices get different colours.Selecting non adjacent squares on a chess boardnumber of combinations colouring 10 eggs with 4 colours if one or 2 colours can be used at the same timeWays to choose balls such that identical colours are adjacent.Given a grid of $m times n$ squares in black and white. Given a rule, which grid sizes $m times n$ (with $m,n ge 3$) have a valid colouring?
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Chess Board (5×5) problem
calculating the number of possible arrangements in chesscombinatorics board with digits neat problemFinding the number of counters having each colour.Board with black and white tilesNumber of colourings for which the vertices of a square can be coloured by three different colours such that adjacent vertices get different colours.Selecting non adjacent squares on a chess boardnumber of combinations colouring 10 eggs with 4 colours if one or 2 colours can be used at the same timeWays to choose balls such that identical colours are adjacent.Given a grid of $m times n$ squares in black and white. Given a rule, which grid sizes $m times n$ (with $m,n ge 3$) have a valid colouring?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty
margin-bottom:0;
.everyonelovesstackoverflowposition:absolute;height:1px;width:1px;opacity:0;top:0;left:0;pointer-events:none;
$begingroup$
25 small squares of a 5×5 chess board are coloured with 5 different
colours available, such that each row contains all 5 available colours
and no two adjacent squares have same colour. Then the no. of
different arrangements possible are?
My attempt:
Let the colours be R,B,G,W,V
To fill the first row I have 5×4×3×2×1
Now for the second row I am sort of messing up I am able to determine 4 cases only but I am getting it at all messed up again and again.
Help me with the editing as well please.
combinatorics
asked Oct 2 at 7:50
DareDevilDareDevil
1815 bronze badges
$endgroup$
add a comment
|
$begingroup$
25 small squares of a 5×5 chess board are coloured with 5 different
colours available, such that each row contains all 5 available colours
and no two adjacent squares have same colour. Then the no. of
different arrangements possible are?
My attempt:
Let the colours be R,B,G,W,V
To fill the first row I have 5×4×3×2×1
Now for the second row I am sort of messing up I am able to determine 4 cases only but I am getting it at all messed up again and again.
Help me with the editing as well please.
combinatorics
asked Oct 2 at 7:50
DareDevilDareDevil
1815 bronze badges
$endgroup$
1
$begingroup$
Each row is a derangement of the previous row. There are $44$ derangements of 5 items. If you think of it as a permutation of the row, the only kinds of permutation of 5 items where everything moves are a 5-cycle (there are $4!$ of those) or a 2-cycle combined with a 3-cycle (there are $binom52 * 2$ of those).
$endgroup$
– Jaap Scherphuis
Oct 2 at 8:05
$begingroup$
Thank you that makes sense.
$endgroup$
– DareDevil
Oct 2 at 8:16
add a comment
|
$begingroup$
25 small squares of a 5×5 chess board are coloured with 5 different
colours available, such that each row contains all 5 available colours
and no two adjacent squares have same colour. Then the no. of
different arrangements possible are?
My attempt:
Let the colours be R,B,G,W,V
To fill the first row I have 5×4×3×2×1
Now for the second row I am sort of messing up I am able to determine 4 cases only but I am getting it at all messed up again and again.
Help me with the editing as well please.
combinatorics
asked Oct 2 at 7:50
DareDevilDareDevil
1815 bronze badges
$endgroup$
25 small squares of a 5×5 chess board are coloured with 5 different
colours available, such that each row contains all 5 available colours
and no two adjacent squares have same colour. Then the no. of
different arrangements possible are?
My attempt:
Let the colours be R,B,G,W,V
To fill the first row I have 5×4×3×2×1
Now for the second row I am sort of messing up I am able to determine 4 cases only but I am getting it at all messed up again and again.
Help me with the editing as well please.
combinatorics
combinatorics
asked Oct 2 at 7:50
DareDevilDareDevil
1815 bronze badges
asked Oct 2 at 7:50
DareDevilDareDevil
1815 bronze badges
asked Oct 2 at 7:50
DareDevilDareDevil
1815 bronze badges
asked Oct 2 at 7:50
DareDevilDareDevil
1815 bronze badges
asked Oct 2 at 7:50
DareDevilDareDevil
1815 bronze badges
1815 bronze badges
1
$begingroup$
Each row is a derangement of the previous row. There are $44$ derangements of 5 items. If you think of it as a permutation of the row, the only kinds of permutation of 5 items where everything moves are a 5-cycle (there are $4!$ of those) or a 2-cycle combined with a 3-cycle (there are $binom52 * 2$ of those).
$endgroup$
– Jaap Scherphuis
Oct 2 at 8:05
$begingroup$
Thank you that makes sense.
$endgroup$
– DareDevil
Oct 2 at 8:16
add a comment
|
1
$begingroup$
Each row is a derangement of the previous row. There are $44$ derangements of 5 items. If you think of it as a permutation of the row, the only kinds of permutation of 5 items where everything moves are a 5-cycle (there are $4!$ of those) or a 2-cycle combined with a 3-cycle (there are $binom52 * 2$ of those).
$endgroup$
– Jaap Scherphuis
Oct 2 at 8:05
$begingroup$
Thank you that makes sense.
$endgroup$
– DareDevil
Oct 2 at 8:16
1
1
$begingroup$
Each row is a derangement of the previous row. There are $44$ derangements of 5 items. If you think of it as a permutation of the row, the only kinds of permutation of 5 items where everything moves are a 5-cycle (there are $4!$ of those) or a 2-cycle combined with a 3-cycle (there are $binom52 * 2$ of those).
$endgroup$
– Jaap Scherphuis
Oct 2 at 8:05
$begingroup$
Each row is a derangement of the previous row. There are $44$ derangements of 5 items. If you think of it as a permutation of the row, the only kinds of permutation of 5 items where everything moves are a 5-cycle (there are $4!$ of those) or a 2-cycle combined with a 3-cycle (there are $binom52 * 2$ of those).
$endgroup$
– Jaap Scherphuis
Oct 2 at 8:05
$begingroup$
Thank you that makes sense.
$endgroup$
– DareDevil
Oct 2 at 8:16
$begingroup$
Thank you that makes sense.
$endgroup$
– DareDevil
Oct 2 at 8:16
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
Consider the following chess table:
$$beginarray hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Let's fill the first row with any colors:
$$beginarray hline a& b& c& d& e \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
The ways to choose $(a,b,c,d,e)$ is $5!$
For the second row, apply inclusion-exclusion.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Total possible concievable arrangements of $(a',b',c',d',e')$ are again $5!$ but we must remove some unfavorable cases.
This is known as a derangement. So, we have counted in our $5!$ some cases where one of $l = l'$ where $l$ is one of $a,b,c,d,e$. The number of such cases is $binom 514!$. But, in this, we have removed some cases where $l=l'$ for two $l$ twice (once for one $l$, another time for the other). The number of such cases is $binom 523!$. Continuing like this, we have the total good arrangements as $5!left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)$. Once we have set the arrangement for this row, we may repeat the process for the next row, and so on.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline a''&b'' &c'' &d'' &e'' \ hline & & & & \ hline & & & & \ hline endarray$$
So, by my calculation, the total favorable arrangements is $$5!^5left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)^4$$
$$= 120*44^4boxed=449771520$$
answered Oct 2 at 8:16
Certainly not a dogCertainly not a dog
1,2432 silver badges15 bronze badges
$endgroup$
1
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil Cheers, glad to help.
$endgroup$
– Certainly not a dog
Oct 2 at 13:51
add a comment
|
$begingroup$
As you noticed for the first we have $5!$ arrangement, for the second row we can use inclusion and exclusion principle as follows.
Notably, the number of cases for the second row with at least two adjacent squares with the same colour with respect to the first row, by inclusion and exclusion principle, is:
$$5cdot 4!-binom52cdot 3!+binom53cdot 2!-binom54cdot 1!+binom55cdot 0!=76$$
therefore the number of cases for the second row with no adjacent squares with the same colour with respect to the first row is:
$$5!-76=120-76=44$$
which is valid for all the others three rows and then finally, by multiplication rule, we obtain:
$$5!cdot (44)^4$$
answered Oct 2 at 9:18
useruser
112k10 gold badges50 silver badges102 bronze badges
$endgroup$
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil You are welcome! Bye
$endgroup$
– user
Oct 2 at 12:30
add a comment
|
Your Answer
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$begingroup$
Consider the following chess table:
$$beginarray hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Let's fill the first row with any colors:
$$beginarray hline a& b& c& d& e \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
The ways to choose $(a,b,c,d,e)$ is $5!$
For the second row, apply inclusion-exclusion.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Total possible concievable arrangements of $(a',b',c',d',e')$ are again $5!$ but we must remove some unfavorable cases.
This is known as a derangement. So, we have counted in our $5!$ some cases where one of $l = l'$ where $l$ is one of $a,b,c,d,e$. The number of such cases is $binom 514!$. But, in this, we have removed some cases where $l=l'$ for two $l$ twice (once for one $l$, another time for the other). The number of such cases is $binom 523!$. Continuing like this, we have the total good arrangements as $5!left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)$. Once we have set the arrangement for this row, we may repeat the process for the next row, and so on.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline a''&b'' &c'' &d'' &e'' \ hline & & & & \ hline & & & & \ hline endarray$$
So, by my calculation, the total favorable arrangements is $$5!^5left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)^4$$
$$= 120*44^4boxed=449771520$$
answered Oct 2 at 8:16
Certainly not a dogCertainly not a dog
1,2432 silver badges15 bronze badges
$endgroup$
1
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil Cheers, glad to help.
$endgroup$
– Certainly not a dog
Oct 2 at 13:51
add a comment
|
$begingroup$
Consider the following chess table:
$$beginarray hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Let's fill the first row with any colors:
$$beginarray hline a& b& c& d& e \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
The ways to choose $(a,b,c,d,e)$ is $5!$
For the second row, apply inclusion-exclusion.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Total possible concievable arrangements of $(a',b',c',d',e')$ are again $5!$ but we must remove some unfavorable cases.
This is known as a derangement. So, we have counted in our $5!$ some cases where one of $l = l'$ where $l$ is one of $a,b,c,d,e$. The number of such cases is $binom 514!$. But, in this, we have removed some cases where $l=l'$ for two $l$ twice (once for one $l$, another time for the other). The number of such cases is $binom 523!$. Continuing like this, we have the total good arrangements as $5!left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)$. Once we have set the arrangement for this row, we may repeat the process for the next row, and so on.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline a''&b'' &c'' &d'' &e'' \ hline & & & & \ hline & & & & \ hline endarray$$
So, by my calculation, the total favorable arrangements is $$5!^5left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)^4$$
$$= 120*44^4boxed=449771520$$
answered Oct 2 at 8:16
Certainly not a dogCertainly not a dog
1,2432 silver badges15 bronze badges
$endgroup$
1
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil Cheers, glad to help.
$endgroup$
– Certainly not a dog
Oct 2 at 13:51
add a comment
|
$begingroup$
Consider the following chess table:
$$beginarray hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Let's fill the first row with any colors:
$$beginarray hline a& b& c& d& e \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
The ways to choose $(a,b,c,d,e)$ is $5!$
For the second row, apply inclusion-exclusion.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Total possible concievable arrangements of $(a',b',c',d',e')$ are again $5!$ but we must remove some unfavorable cases.
This is known as a derangement. So, we have counted in our $5!$ some cases where one of $l = l'$ where $l$ is one of $a,b,c,d,e$. The number of such cases is $binom 514!$. But, in this, we have removed some cases where $l=l'$ for two $l$ twice (once for one $l$, another time for the other). The number of such cases is $binom 523!$. Continuing like this, we have the total good arrangements as $5!left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)$. Once we have set the arrangement for this row, we may repeat the process for the next row, and so on.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline a''&b'' &c'' &d'' &e'' \ hline & & & & \ hline & & & & \ hline endarray$$
So, by my calculation, the total favorable arrangements is $$5!^5left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)^4$$
$$= 120*44^4boxed=449771520$$
answered Oct 2 at 8:16
Certainly not a dogCertainly not a dog
1,2432 silver badges15 bronze badges
$endgroup$
Consider the following chess table:
$$beginarray hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Let's fill the first row with any colors:
$$beginarray hline a& b& c& d& e \ hline & & & & \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
The ways to choose $(a,b,c,d,e)$ is $5!$
For the second row, apply inclusion-exclusion.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline & & & & \ hline & & & & \ hline & & & & \ hline endarray$$
Total possible concievable arrangements of $(a',b',c',d',e')$ are again $5!$ but we must remove some unfavorable cases.
This is known as a derangement. So, we have counted in our $5!$ some cases where one of $l = l'$ where $l$ is one of $a,b,c,d,e$. The number of such cases is $binom 514!$. But, in this, we have removed some cases where $l=l'$ for two $l$ twice (once for one $l$, another time for the other). The number of such cases is $binom 523!$. Continuing like this, we have the total good arrangements as $5!left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)$. Once we have set the arrangement for this row, we may repeat the process for the next row, and so on.
$$beginarray hline a& b& c& d&e \ hline a'&b' &c' &d' &e' \ hline a''&b'' &c'' &d'' &e'' \ hline & & & & \ hline & & & & \ hline endarray$$
So, by my calculation, the total favorable arrangements is $$5!^5left(displaystylesum_k = 0^5left(-1right)^kfrac 1k!right)^4$$
$$= 120*44^4boxed=449771520$$
answered Oct 2 at 8:16
Certainly not a dogCertainly not a dog
1,2432 silver badges15 bronze badges
edited Oct 2 at 9:11
answered Oct 2 at 8:16
Certainly not a dogCertainly not a dog
1,2432 silver badges15 bronze badges
answered Oct 2 at 8:16
Certainly not a dogCertainly not a dog
1,2432 silver badges15 bronze badges
answered Oct 2 at 8:16
Certainly not a dogCertainly not a dog
1,2432 silver badges15 bronze badges
1,2432 silver badges15 bronze badges
1
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil Cheers, glad to help.
$endgroup$
– Certainly not a dog
Oct 2 at 13:51
add a comment
|
1
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil Cheers, glad to help.
$endgroup$
– Certainly not a dog
Oct 2 at 13:51
1
1
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil Cheers, glad to help.
$endgroup$
– Certainly not a dog
Oct 2 at 13:51
$begingroup$
@DareDevil Cheers, glad to help.
$endgroup$
– Certainly not a dog
Oct 2 at 13:51
add a comment
|
$begingroup$
As you noticed for the first we have $5!$ arrangement, for the second row we can use inclusion and exclusion principle as follows.
Notably, the number of cases for the second row with at least two adjacent squares with the same colour with respect to the first row, by inclusion and exclusion principle, is:
$$5cdot 4!-binom52cdot 3!+binom53cdot 2!-binom54cdot 1!+binom55cdot 0!=76$$
therefore the number of cases for the second row with no adjacent squares with the same colour with respect to the first row is:
$$5!-76=120-76=44$$
which is valid for all the others three rows and then finally, by multiplication rule, we obtain:
$$5!cdot (44)^4$$
answered Oct 2 at 9:18
useruser
112k10 gold badges50 silver badges102 bronze badges
$endgroup$
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil You are welcome! Bye
$endgroup$
– user
Oct 2 at 12:30
add a comment
|
$begingroup$
As you noticed for the first we have $5!$ arrangement, for the second row we can use inclusion and exclusion principle as follows.
Notably, the number of cases for the second row with at least two adjacent squares with the same colour with respect to the first row, by inclusion and exclusion principle, is:
$$5cdot 4!-binom52cdot 3!+binom53cdot 2!-binom54cdot 1!+binom55cdot 0!=76$$
therefore the number of cases for the second row with no adjacent squares with the same colour with respect to the first row is:
$$5!-76=120-76=44$$
which is valid for all the others three rows and then finally, by multiplication rule, we obtain:
$$5!cdot (44)^4$$
answered Oct 2 at 9:18
useruser
112k10 gold badges50 silver badges102 bronze badges
$endgroup$
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil You are welcome! Bye
$endgroup$
– user
Oct 2 at 12:30
add a comment
|
$begingroup$
As you noticed for the first we have $5!$ arrangement, for the second row we can use inclusion and exclusion principle as follows.
Notably, the number of cases for the second row with at least two adjacent squares with the same colour with respect to the first row, by inclusion and exclusion principle, is:
$$5cdot 4!-binom52cdot 3!+binom53cdot 2!-binom54cdot 1!+binom55cdot 0!=76$$
therefore the number of cases for the second row with no adjacent squares with the same colour with respect to the first row is:
$$5!-76=120-76=44$$
which is valid for all the others three rows and then finally, by multiplication rule, we obtain:
$$5!cdot (44)^4$$
answered Oct 2 at 9:18
useruser
112k10 gold badges50 silver badges102 bronze badges
$endgroup$
As you noticed for the first we have $5!$ arrangement, for the second row we can use inclusion and exclusion principle as follows.
Notably, the number of cases for the second row with at least two adjacent squares with the same colour with respect to the first row, by inclusion and exclusion principle, is:
$$5cdot 4!-binom52cdot 3!+binom53cdot 2!-binom54cdot 1!+binom55cdot 0!=76$$
therefore the number of cases for the second row with no adjacent squares with the same colour with respect to the first row is:
$$5!-76=120-76=44$$
which is valid for all the others three rows and then finally, by multiplication rule, we obtain:
$$5!cdot (44)^4$$
answered Oct 2 at 9:18
useruser
112k10 gold badges50 silver badges102 bronze badges
answered Oct 2 at 9:18
useruser
112k10 gold badges50 silver badges102 bronze badges
answered Oct 2 at 9:18
useruser
112k10 gold badges50 silver badges102 bronze badges
answered Oct 2 at 9:18
useruser
112k10 gold badges50 silver badges102 bronze badges
112k10 gold badges50 silver badges102 bronze badges
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil You are welcome! Bye
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– user
Oct 2 at 12:30
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Thank you for your efforts.
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– DareDevil
Oct 2 at 12:28
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@DareDevil You are welcome! Bye
$endgroup$
– user
Oct 2 at 12:30
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Thank you for your efforts.
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– DareDevil
Oct 2 at 12:28
$begingroup$
Thank you for your efforts.
$endgroup$
– DareDevil
Oct 2 at 12:28
$begingroup$
@DareDevil You are welcome! Bye
$endgroup$
– user
Oct 2 at 12:30
$begingroup$
@DareDevil You are welcome! Bye
$endgroup$
– user
Oct 2 at 12:30
add a comment
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Each row is a derangement of the previous row. There are $44$ derangements of 5 items. If you think of it as a permutation of the row, the only kinds of permutation of 5 items where everything moves are a 5-cycle (there are $4!$ of those) or a 2-cycle combined with a 3-cycle (there are $binom52 * 2$ of those).
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– Jaap Scherphuis
Oct 2 at 8:05
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Thank you that makes sense.
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– DareDevil
Oct 2 at 8:16