How to calculate a subgroup of some permutation?
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
3,360320
asked Dec 1 at 3:32
Enrique
1032
add a comment |
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
3,360320
asked Dec 1 at 3:32
Enrique
1032
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 |
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
3,360320
asked Dec 1 at 3:32
Enrique
1032
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
3,360320
asked Dec 1 at 3:32
Enrique
1032
edited Dec 1 at 3:45
platty
3,360320
asked Dec 1 at 3:32
Enrique
1032
edited Dec 1 at 3:45
platty
3,360320
edited Dec 1 at 3:45
platty
3,360320
edited Dec 1 at 3:45
platty
3,360320
3,360320
asked Dec 1 at 3:32
Enrique
1032
asked Dec 1 at 3:32
Enrique
1032
asked Dec 1 at 3:32
Enrique
1032
1032
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
562
add a comment |
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1 Answer
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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
562
add a comment |
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
562
add a comment |
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
562
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
562
answered Dec 1 at 3:54
eKoontz
562
answered Dec 1 at 3:54
eKoontz
562
answered Dec 1 at 3:54
eKoontz
562
562
add a comment |
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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