How to calculate a subgroup of some permutation?
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0
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I think is easier to explain it with a specific problem (but would be nice a generic answer):
We have 2 options, $A$ and $B$, counting permutations of $N$ items is easy, but how to calculate the subgroup where we have two $A$'s?
Example with $N = 3$:
$AAA$
$AAB$
$ABA$
$BAA$
$ABB$
$BBA$
$BAB$
$BBB$
To get the $8$ options is just $2^3$, but what is the formula to get "$3$" (permutation with 2 $A$)?
combinatorics permutations
edited Dec 1 at 3:45
platty
2,612218
asked Dec 1 at 3:32
Enrique
1011
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Enrique is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
0
down vote
favorite
I think is easier to explain it with a specific problem (but would be nice a generic answer):
We have 2 options, $A$ and $B$, counting permutations of $N$ items is easy, but how to calculate the subgroup where we have two $A$'s?
Example with $N = 3$:
$AAA$
$AAB$
$ABA$
$BAA$
$ABB$
$BBA$
$BAB$
$BBB$
To get the $8$ options is just $2^3$, but what is the formula to get "$3$" (permutation with 2 $A$)?
combinatorics permutations
edited Dec 1 at 3:45
platty
2,612218
asked Dec 1 at 3:32
Enrique
1011
New contributor
Enrique is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– platty
Dec 1 at 3:45
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I think is easier to explain it with a specific problem (but would be nice a generic answer):
We have 2 options, $A$ and $B$, counting permutations of $N$ items is easy, but how to calculate the subgroup where we have two $A$'s?
Example with $N = 3$:
$AAA$
$AAB$
$ABA$
$BAA$
$ABB$
$BBA$
$BAB$
$BBB$
To get the $8$ options is just $2^3$, but what is the formula to get "$3$" (permutation with 2 $A$)?
combinatorics permutations
edited Dec 1 at 3:45
platty
2,612218
asked Dec 1 at 3:32
Enrique
1011
New contributor
Enrique is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I think is easier to explain it with a specific problem (but would be nice a generic answer):
We have 2 options, $A$ and $B$, counting permutations of $N$ items is easy, but how to calculate the subgroup where we have two $A$'s?
Example with $N = 3$:
$AAA$
$AAB$
$ABA$
$BAA$
$ABB$
$BBA$
$BAB$
$BBB$
To get the $8$ options is just $2^3$, but what is the formula to get "$3$" (permutation with 2 $A$)?
combinatorics permutations
combinatorics permutations
edited Dec 1 at 3:45
platty
2,612218
asked Dec 1 at 3:32
Enrique
1011
New contributor
Enrique is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 1 at 3:45
platty
2,612218
asked Dec 1 at 3:32
Enrique
1011
New contributor
Enrique is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 1 at 3:45
platty
2,612218
edited Dec 1 at 3:45
platty
2,612218
edited Dec 1 at 3:45
platty
2,612218
2,612218
asked Dec 1 at 3:32
Enrique
1011
New contributor
Enrique is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 1 at 3:32
Enrique
1011
asked Dec 1 at 3:32
Enrique
1011
1011
New contributor
Enrique is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Enrique is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Enrique is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– platty
Dec 1 at 3:45
add a comment |
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– platty
Dec 1 at 3:45
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– platty
Dec 1 at 3:45
Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– platty
Dec 1 at 3:45
add a comment |
1 Answer
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I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.
answered Dec 1 at 3:54
eKoontz
412
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.
answered Dec 1 at 3:54
eKoontz
412
New contributor
eKoontz 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 |
up vote
4
down vote
I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.
answered Dec 1 at 3:54
eKoontz
412
New contributor
eKoontz 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 |
up vote
4
down vote
up vote
4
down vote
I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.
answered Dec 1 at 3:54
eKoontz
412
New contributor
eKoontz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I might be oversimplifying this, but if order does not matter, then it's simply a question of how many ways can you place two A's in a list of three, which is ${3 choose 2} = 3.$ In general, for N objects with k number of A's it'd be ${N choose k} = frac{N!}{k!(N-k)!}$.
answered Dec 1 at 3:54
eKoontz
412
New contributor
eKoontz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 1 at 3:54
eKoontz
412
New contributor
eKoontz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Dec 1 at 3:54
eKoontz
412
answered Dec 1 at 3:54
eKoontz
412
412
New contributor
eKoontz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
eKoontz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
eKoontz 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 |
Enrique is a new contributor. Be nice, and check out our Code of Conduct.
Enrique is a new contributor. Be nice, and check out our Code of Conduct.
Enrique is a new contributor. Be nice, and check out our Code of Conduct.
Enrique is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to MSE! For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– platty
Dec 1 at 3:45