Permutation: How to arrange 12 people around a table for 7?possible seeting arrangments given number of people, and number of chairs, chairs can be emptyHow many ways can seven people sit around a circular table?How many ways can this be done?Combinatorics - seating 7 people around table with 8 seats; two people have to be two seats apartPlacing 8 people around a table so that 2 people never sit togetherSeating people at a table, but some don't like each otherFind the position of ball on the $2017^th$ stepCount the possible ways to seat people at a round tableBoy Girl round table permutation problemWays to arrange $ngeq2$ people around a circular table, given two permanent seats.
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Permutation: How to arrange 12 people around a table for 7?
possible seeting arrangments given number of people, and number of chairs, chairs can be emptyHow many ways can seven people sit around a circular table?How many ways can this be done?Combinatorics - seating 7 people around table with 8 seats; two people have to be two seats apartPlacing 8 people around a table so that 2 people never sit togetherSeating people at a table, but some don't like each otherFind the position of ball on the $2017^th$ stepCount the possible ways to seat people at a round tableBoy Girl round table permutation problemWays to arrange $ngeq2$ people around a circular table, given two permanent seats.
$begingroup$
I want to understand how to arrange $12$ people around a circular table with $7$ chairs. We don't care about the overflow, those people can go to another table.
I thought the way to solve the problem is that the position for the first chair is fixed, the second chair has $11$ possible options of people (since one person is already seated), the third chair has $10$ possible options, the fourth chair has $9$ possible options and so on until we get to the seventh chair which has $6$ possible options of people.
So I thought the way to solve is that this is a permutation problem $1*11*10*9*8*7*6=332640=11P6=frac12P712$. But my professor says the correct answer is $frac12P77$. I don't understand why we should divide $12P7$ by the number of chairs. Can someone explain this me?
combinatorics permutations binomial-coefficients factorial
edited 2 days ago
Michael Rozenberg
108k1895200
asked Mar 10 at 20:02
SamSam
4048
$endgroup$
add a comment |
$begingroup$
I want to understand how to arrange $12$ people around a circular table with $7$ chairs. We don't care about the overflow, those people can go to another table.
I thought the way to solve the problem is that the position for the first chair is fixed, the second chair has $11$ possible options of people (since one person is already seated), the third chair has $10$ possible options, the fourth chair has $9$ possible options and so on until we get to the seventh chair which has $6$ possible options of people.
So I thought the way to solve is that this is a permutation problem $1*11*10*9*8*7*6=332640=11P6=frac12P712$. But my professor says the correct answer is $frac12P77$. I don't understand why we should divide $12P7$ by the number of chairs. Can someone explain this me?
combinatorics permutations binomial-coefficients factorial
edited 2 days ago
Michael Rozenberg
108k1895200
asked Mar 10 at 20:02
SamSam
4048
$endgroup$
2
$begingroup$
Choose which seven people get a seat. Then, let the youngest of those people sit down first at the table wherever they like. Then, fill the remaining six of the seven chosen people around the table.
$endgroup$
– JMoravitz
Mar 10 at 20:07
add a comment |
$begingroup$
I want to understand how to arrange $12$ people around a circular table with $7$ chairs. We don't care about the overflow, those people can go to another table.
I thought the way to solve the problem is that the position for the first chair is fixed, the second chair has $11$ possible options of people (since one person is already seated), the third chair has $10$ possible options, the fourth chair has $9$ possible options and so on until we get to the seventh chair which has $6$ possible options of people.
So I thought the way to solve is that this is a permutation problem $1*11*10*9*8*7*6=332640=11P6=frac12P712$. But my professor says the correct answer is $frac12P77$. I don't understand why we should divide $12P7$ by the number of chairs. Can someone explain this me?
combinatorics permutations binomial-coefficients factorial
edited 2 days ago
Michael Rozenberg
108k1895200
asked Mar 10 at 20:02
SamSam
4048
$endgroup$
I want to understand how to arrange $12$ people around a circular table with $7$ chairs. We don't care about the overflow, those people can go to another table.
I thought the way to solve the problem is that the position for the first chair is fixed, the second chair has $11$ possible options of people (since one person is already seated), the third chair has $10$ possible options, the fourth chair has $9$ possible options and so on until we get to the seventh chair which has $6$ possible options of people.
So I thought the way to solve is that this is a permutation problem $1*11*10*9*8*7*6=332640=11P6=frac12P712$. But my professor says the correct answer is $frac12P77$. I don't understand why we should divide $12P7$ by the number of chairs. Can someone explain this me?
combinatorics permutations binomial-coefficients factorial
combinatorics permutations binomial-coefficients factorial
edited 2 days ago
Michael Rozenberg
108k1895200
asked Mar 10 at 20:02
SamSam
4048
edited 2 days ago
Michael Rozenberg
108k1895200
asked Mar 10 at 20:02
SamSam
4048
edited 2 days ago
Michael Rozenberg
108k1895200
edited 2 days ago
Michael Rozenberg
108k1895200
edited 2 days ago
Michael Rozenberg
108k1895200
108k1895200
asked Mar 10 at 20:02
SamSam
4048
asked Mar 10 at 20:02
SamSam
4048
asked Mar 10 at 20:02
SamSam
4048
4048
2
$begingroup$
Choose which seven people get a seat. Then, let the youngest of those people sit down first at the table wherever they like. Then, fill the remaining six of the seven chosen people around the table.
$endgroup$
– JMoravitz
Mar 10 at 20:07
add a comment |
2
$begingroup$
Choose which seven people get a seat. Then, let the youngest of those people sit down first at the table wherever they like. Then, fill the remaining six of the seven chosen people around the table.
$endgroup$
– JMoravitz
Mar 10 at 20:07
2
2
$begingroup$
Choose which seven people get a seat. Then, let the youngest of those people sit down first at the table wherever they like. Then, fill the remaining six of the seven chosen people around the table.
$endgroup$
– JMoravitz
Mar 10 at 20:07
$begingroup$
Choose which seven people get a seat. Then, let the youngest of those people sit down first at the table wherever they like. Then, fill the remaining six of the seven chosen people around the table.
$endgroup$
– JMoravitz
Mar 10 at 20:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think it should be $$binom1276!$$
answered Mar 10 at 20:11
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Why multiply by $6!$?
$endgroup$
– Sam
Mar 10 at 20:14
1
$begingroup$
@Sam Because we take one of these seven and for the rest we make all permutations. We think, of course, that any cyclic permutations of these seven are the same.
$endgroup$
– Michael Rozenberg
Mar 10 at 20:16
1
$begingroup$
@Sam Note that $binom1276!$ is equal to $frac_12P_77$. I much prefer this answer (which matches mine above in the comments) as it avoids the "division by symmetry" style arguments that seem common and confuses people.
$endgroup$
– JMoravitz
Mar 10 at 20:30
add a comment |
$begingroup$
If the only important thing while arranging places is the locations of people relative to each other; or in other word, if nothing changes when we rotate all the chairs together around the circular table, according to the problem, professor is right. However, if the positions of chairs are important for us, such as sitting around circular table in a conference (Somebody must be turns behind to the scene), it would not be different from line-permutation .
Dividing by the number of chairs is comes from the rotation. If you rotate all chairs together 1,2,3,...,n times, it does not change location of people relative to each other, but it does relative to room, scene, board etc.
answered Mar 10 at 20:31
Ebubekir İnalEbubekir İnal
1
New contributor
Ebubekir İnal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think it should be $$binom1276!$$
answered Mar 10 at 20:11
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Why multiply by $6!$?
$endgroup$
– Sam
Mar 10 at 20:14
1
$begingroup$
@Sam Because we take one of these seven and for the rest we make all permutations. We think, of course, that any cyclic permutations of these seven are the same.
$endgroup$
– Michael Rozenberg
Mar 10 at 20:16
1
$begingroup$
@Sam Note that $binom1276!$ is equal to $frac_12P_77$. I much prefer this answer (which matches mine above in the comments) as it avoids the "division by symmetry" style arguments that seem common and confuses people.
$endgroup$
– JMoravitz
Mar 10 at 20:30
add a comment |
$begingroup$
I think it should be $$binom1276!$$
answered Mar 10 at 20:11
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
$begingroup$
Why multiply by $6!$?
$endgroup$
– Sam
Mar 10 at 20:14
1
$begingroup$
@Sam Because we take one of these seven and for the rest we make all permutations. We think, of course, that any cyclic permutations of these seven are the same.
$endgroup$
– Michael Rozenberg
Mar 10 at 20:16
1
$begingroup$
@Sam Note that $binom1276!$ is equal to $frac_12P_77$. I much prefer this answer (which matches mine above in the comments) as it avoids the "division by symmetry" style arguments that seem common and confuses people.
$endgroup$
– JMoravitz
Mar 10 at 20:30
add a comment |
$begingroup$
I think it should be $$binom1276!$$
answered Mar 10 at 20:11
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
I think it should be $$binom1276!$$
answered Mar 10 at 20:11
Michael RozenbergMichael Rozenberg
108k1895200
answered Mar 10 at 20:11
Michael RozenbergMichael Rozenberg
108k1895200
answered Mar 10 at 20:11
Michael RozenbergMichael Rozenberg
108k1895200
answered Mar 10 at 20:11
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
$begingroup$
Why multiply by $6!$?
$endgroup$
– Sam
Mar 10 at 20:14
1
$begingroup$
@Sam Because we take one of these seven and for the rest we make all permutations. We think, of course, that any cyclic permutations of these seven are the same.
$endgroup$
– Michael Rozenberg
Mar 10 at 20:16
1
$begingroup$
@Sam Note that $binom1276!$ is equal to $frac_12P_77$. I much prefer this answer (which matches mine above in the comments) as it avoids the "division by symmetry" style arguments that seem common and confuses people.
$endgroup$
– JMoravitz
Mar 10 at 20:30
add a comment |
$begingroup$
Why multiply by $6!$?
$endgroup$
– Sam
Mar 10 at 20:14
1
$begingroup$
@Sam Because we take one of these seven and for the rest we make all permutations. We think, of course, that any cyclic permutations of these seven are the same.
$endgroup$
– Michael Rozenberg
Mar 10 at 20:16
1
$begingroup$
@Sam Note that $binom1276!$ is equal to $frac_12P_77$. I much prefer this answer (which matches mine above in the comments) as it avoids the "division by symmetry" style arguments that seem common and confuses people.
$endgroup$
– JMoravitz
Mar 10 at 20:30
$begingroup$
Why multiply by $6!$?
$endgroup$
– Sam
Mar 10 at 20:14
$begingroup$
Why multiply by $6!$?
$endgroup$
– Sam
Mar 10 at 20:14
1
1
$begingroup$
@Sam Because we take one of these seven and for the rest we make all permutations. We think, of course, that any cyclic permutations of these seven are the same.
$endgroup$
– Michael Rozenberg
Mar 10 at 20:16
$begingroup$
@Sam Because we take one of these seven and for the rest we make all permutations. We think, of course, that any cyclic permutations of these seven are the same.
$endgroup$
– Michael Rozenberg
Mar 10 at 20:16
1
1
$begingroup$
@Sam Note that $binom1276!$ is equal to $frac_12P_77$. I much prefer this answer (which matches mine above in the comments) as it avoids the "division by symmetry" style arguments that seem common and confuses people.
$endgroup$
– JMoravitz
Mar 10 at 20:30
$begingroup$
@Sam Note that $binom1276!$ is equal to $frac_12P_77$. I much prefer this answer (which matches mine above in the comments) as it avoids the "division by symmetry" style arguments that seem common and confuses people.
$endgroup$
– JMoravitz
Mar 10 at 20:30
add a comment |
$begingroup$
If the only important thing while arranging places is the locations of people relative to each other; or in other word, if nothing changes when we rotate all the chairs together around the circular table, according to the problem, professor is right. However, if the positions of chairs are important for us, such as sitting around circular table in a conference (Somebody must be turns behind to the scene), it would not be different from line-permutation .
Dividing by the number of chairs is comes from the rotation. If you rotate all chairs together 1,2,3,...,n times, it does not change location of people relative to each other, but it does relative to room, scene, board etc.
answered Mar 10 at 20:31
Ebubekir İnalEbubekir İnal
1
New contributor
Ebubekir İnal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If the only important thing while arranging places is the locations of people relative to each other; or in other word, if nothing changes when we rotate all the chairs together around the circular table, according to the problem, professor is right. However, if the positions of chairs are important for us, such as sitting around circular table in a conference (Somebody must be turns behind to the scene), it would not be different from line-permutation .
Dividing by the number of chairs is comes from the rotation. If you rotate all chairs together 1,2,3,...,n times, it does not change location of people relative to each other, but it does relative to room, scene, board etc.
answered Mar 10 at 20:31
Ebubekir İnalEbubekir İnal
1
New contributor
Ebubekir İnal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
If the only important thing while arranging places is the locations of people relative to each other; or in other word, if nothing changes when we rotate all the chairs together around the circular table, according to the problem, professor is right. However, if the positions of chairs are important for us, such as sitting around circular table in a conference (Somebody must be turns behind to the scene), it would not be different from line-permutation .
Dividing by the number of chairs is comes from the rotation. If you rotate all chairs together 1,2,3,...,n times, it does not change location of people relative to each other, but it does relative to room, scene, board etc.
answered Mar 10 at 20:31
Ebubekir İnalEbubekir İnal
1
New contributor
Ebubekir İnal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
If the only important thing while arranging places is the locations of people relative to each other; or in other word, if nothing changes when we rotate all the chairs together around the circular table, according to the problem, professor is right. However, if the positions of chairs are important for us, such as sitting around circular table in a conference (Somebody must be turns behind to the scene), it would not be different from line-permutation .
Dividing by the number of chairs is comes from the rotation. If you rotate all chairs together 1,2,3,...,n times, it does not change location of people relative to each other, but it does relative to room, scene, board etc.
answered Mar 10 at 20:31
Ebubekir İnalEbubekir İnal
1
New contributor
Ebubekir İnal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 10 at 20:31
Ebubekir İnalEbubekir İnal
1
New contributor
Ebubekir İnal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 10 at 20:31
Ebubekir İnalEbubekir İnal
1
answered Mar 10 at 20:31
Ebubekir İnalEbubekir İnal
1
1
New contributor
Ebubekir İnal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ebubekir İnal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ebubekir İnal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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$begingroup$
Choose which seven people get a seat. Then, let the youngest of those people sit down first at the table wherever they like. Then, fill the remaining six of the seven chosen people around the table.
$endgroup$
– JMoravitz
Mar 10 at 20:07