six groups of six
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I was given a problem by a friend, and I'd like some advice on how it can be solved. There are 36 students in all. Each "arrangement" breaks the students down into 6 groups of 6. The aim is to find all the arrangements of these students such that the same student does not ever sit in the same group with another student more than once. I never studied this type of math (I don't think!), so advice for an algorithm would be appreciated!
combinatorics
asked Aug 11 at 0:26
Adam Mac
1
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up vote
-2
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I was given a problem by a friend, and I'd like some advice on how it can be solved. There are 36 students in all. Each "arrangement" breaks the students down into 6 groups of 6. The aim is to find all the arrangements of these students such that the same student does not ever sit in the same group with another student more than once. I never studied this type of math (I don't think!), so advice for an algorithm would be appreciated!
combinatorics
asked Aug 11 at 0:26
Adam Mac
1
Welcome to math.SE! You may find our take on questions to be a tad different than other sites. You'll find questions of the form "Here's a problem" with nothing else tend to be ill-received; it would be preferred that more context be provided and that you avoid no clue questions. For example, have you tried writing down a few examples? (Note that we encourage this not only to help us help you, but it'll likely help you as well.)
– Simply Beautiful Art
Aug 11 at 0:41
Do a first arrangement and number the students $1,2,3,...,36$ such that the groups are constitutive $1,2,3,4,5,6$ the first group, $7,8,9,10,11,12$ the second, etc. For all subsequent arrangements each student from the same group in the initial arrangement must now be in different groups. As a consequence, in the group of student $1$ there must always be a one of the students from the second group of in the first arrangement, i.e. one of $7,8,9,10,11,12$. There are only $6$ possibilities without repeating the chosen student from the second group.
– user583012
Aug 11 at 2:07
Therefore, there cannot be more than $6$ further arrangements after the first. Now it is we just need to show that those $6$ further arrangements can be done.For example, you can list them. It is only $6$ more. So, in total $7$ arrangements are possible.
– user583012
Aug 11 at 2:07
Thank you, @fatherBrown. I think I was over-thinking this one. Was also even wondering if it was a good problem for a recursion algorithm to solve....
– Adam Mac
Aug 11 at 12:31
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I was given a problem by a friend, and I'd like some advice on how it can be solved. There are 36 students in all. Each "arrangement" breaks the students down into 6 groups of 6. The aim is to find all the arrangements of these students such that the same student does not ever sit in the same group with another student more than once. I never studied this type of math (I don't think!), so advice for an algorithm would be appreciated!
combinatorics
asked Aug 11 at 0:26
Adam Mac
1
I was given a problem by a friend, and I'd like some advice on how it can be solved. There are 36 students in all. Each "arrangement" breaks the students down into 6 groups of 6. The aim is to find all the arrangements of these students such that the same student does not ever sit in the same group with another student more than once. I never studied this type of math (I don't think!), so advice for an algorithm would be appreciated!
combinatorics
asked Aug 11 at 0:26
Adam Mac
1
asked Aug 11 at 0:26
Adam Mac
1
asked Aug 11 at 0:26
Adam Mac
1
asked Aug 11 at 0:26
Adam Mac
1
1
Welcome to math.SE! You may find our take on questions to be a tad different than other sites. You'll find questions of the form "Here's a problem" with nothing else tend to be ill-received; it would be preferred that more context be provided and that you avoid no clue questions. For example, have you tried writing down a few examples? (Note that we encourage this not only to help us help you, but it'll likely help you as well.)
– Simply Beautiful Art
Aug 11 at 0:41
Do a first arrangement and number the students $1,2,3,...,36$ such that the groups are constitutive $1,2,3,4,5,6$ the first group, $7,8,9,10,11,12$ the second, etc. For all subsequent arrangements each student from the same group in the initial arrangement must now be in different groups. As a consequence, in the group of student $1$ there must always be a one of the students from the second group of in the first arrangement, i.e. one of $7,8,9,10,11,12$. There are only $6$ possibilities without repeating the chosen student from the second group.
– user583012
Aug 11 at 2:07
Therefore, there cannot be more than $6$ further arrangements after the first. Now it is we just need to show that those $6$ further arrangements can be done.For example, you can list them. It is only $6$ more. So, in total $7$ arrangements are possible.
– user583012
Aug 11 at 2:07
Thank you, @fatherBrown. I think I was over-thinking this one. Was also even wondering if it was a good problem for a recursion algorithm to solve....
– Adam Mac
Aug 11 at 12:31
add a comment |Â
Welcome to math.SE! You may find our take on questions to be a tad different than other sites. You'll find questions of the form "Here's a problem" with nothing else tend to be ill-received; it would be preferred that more context be provided and that you avoid no clue questions. For example, have you tried writing down a few examples? (Note that we encourage this not only to help us help you, but it'll likely help you as well.)
– Simply Beautiful Art
Aug 11 at 0:41
Do a first arrangement and number the students $1,2,3,...,36$ such that the groups are constitutive $1,2,3,4,5,6$ the first group, $7,8,9,10,11,12$ the second, etc. For all subsequent arrangements each student from the same group in the initial arrangement must now be in different groups. As a consequence, in the group of student $1$ there must always be a one of the students from the second group of in the first arrangement, i.e. one of $7,8,9,10,11,12$. There are only $6$ possibilities without repeating the chosen student from the second group.
– user583012
Aug 11 at 2:07
Therefore, there cannot be more than $6$ further arrangements after the first. Now it is we just need to show that those $6$ further arrangements can be done.For example, you can list them. It is only $6$ more. So, in total $7$ arrangements are possible.
– user583012
Aug 11 at 2:07
Thank you, @fatherBrown. I think I was over-thinking this one. Was also even wondering if it was a good problem for a recursion algorithm to solve....
– Adam Mac
Aug 11 at 12:31
Welcome to math.SE! You may find our take on questions to be a tad different than other sites. You'll find questions of the form "Here's a problem" with nothing else tend to be ill-received; it would be preferred that more context be provided and that you avoid no clue questions. For example, have you tried writing down a few examples? (Note that we encourage this not only to help us help you, but it'll likely help you as well.)
– Simply Beautiful Art
Aug 11 at 0:41
Welcome to math.SE! You may find our take on questions to be a tad different than other sites. You'll find questions of the form "Here's a problem" with nothing else tend to be ill-received; it would be preferred that more context be provided and that you avoid no clue questions. For example, have you tried writing down a few examples? (Note that we encourage this not only to help us help you, but it'll likely help you as well.)
– Simply Beautiful Art
Aug 11 at 0:41
Do a first arrangement and number the students $1,2,3,...,36$ such that the groups are constitutive $1,2,3,4,5,6$ the first group, $7,8,9,10,11,12$ the second, etc. For all subsequent arrangements each student from the same group in the initial arrangement must now be in different groups. As a consequence, in the group of student $1$ there must always be a one of the students from the second group of in the first arrangement, i.e. one of $7,8,9,10,11,12$. There are only $6$ possibilities without repeating the chosen student from the second group.
– user583012
Aug 11 at 2:07
Do a first arrangement and number the students $1,2,3,...,36$ such that the groups are constitutive $1,2,3,4,5,6$ the first group, $7,8,9,10,11,12$ the second, etc. For all subsequent arrangements each student from the same group in the initial arrangement must now be in different groups. As a consequence, in the group of student $1$ there must always be a one of the students from the second group of in the first arrangement, i.e. one of $7,8,9,10,11,12$. There are only $6$ possibilities without repeating the chosen student from the second group.
– user583012
Aug 11 at 2:07
Therefore, there cannot be more than $6$ further arrangements after the first. Now it is we just need to show that those $6$ further arrangements can be done.For example, you can list them. It is only $6$ more. So, in total $7$ arrangements are possible.
– user583012
Aug 11 at 2:07
Therefore, there cannot be more than $6$ further arrangements after the first. Now it is we just need to show that those $6$ further arrangements can be done.For example, you can list them. It is only $6$ more. So, in total $7$ arrangements are possible.
– user583012
Aug 11 at 2:07
Thank you, @fatherBrown. I think I was over-thinking this one. Was also even wondering if it was a good problem for a recursion algorithm to solve....
– Adam Mac
Aug 11 at 12:31
Thank you, @fatherBrown. I think I was over-thinking this one. Was also even wondering if it was a good problem for a recursion algorithm to solve....
– Adam Mac
Aug 11 at 12:31
add a comment |Â
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Welcome to math.SE! You may find our take on questions to be a tad different than other sites. You'll find questions of the form "Here's a problem" with nothing else tend to be ill-received; it would be preferred that more context be provided and that you avoid no clue questions. For example, have you tried writing down a few examples? (Note that we encourage this not only to help us help you, but it'll likely help you as well.)
– Simply Beautiful Art
Aug 11 at 0:41
Do a first arrangement and number the students $1,2,3,...,36$ such that the groups are constitutive $1,2,3,4,5,6$ the first group, $7,8,9,10,11,12$ the second, etc. For all subsequent arrangements each student from the same group in the initial arrangement must now be in different groups. As a consequence, in the group of student $1$ there must always be a one of the students from the second group of in the first arrangement, i.e. one of $7,8,9,10,11,12$. There are only $6$ possibilities without repeating the chosen student from the second group.
– user583012
Aug 11 at 2:07
Therefore, there cannot be more than $6$ further arrangements after the first. Now it is we just need to show that those $6$ further arrangements can be done.For example, you can list them. It is only $6$ more. So, in total $7$ arrangements are possible.
– user583012
Aug 11 at 2:07
Thank you, @fatherBrown. I think I was over-thinking this one. Was also even wondering if it was a good problem for a recursion algorithm to solve....
– Adam Mac
Aug 11 at 12:31