For permutations with “copies”, in what sense are the elements of the set distinct?
$begingroup$
Suppose I have a set of elements $$ S = lbrace A,B,B,D,E rbrace$$
Then the number of distinct permutations of size 5 is
$$frac{5!}{2!} = 60$$
This is because $S$ contains a ‘copy’ of B.
But by definition, in a set of elements, each element is distinct. However, now I’m essentially saying “wait actually B and B aren’t distinct.”
So why isn’t the set $$ S = lbrace A,B,D,E rbrace$$ if the B elements are not distinct? Notice the permutations of this set are very different
$$4! = 24$$
Are there two different kinds of being distinct here?
combinatorics
asked Dec 17 '18 at 22:29
Stan ShunpikeStan Shunpike
1,83111439
$endgroup$
add a comment |
$begingroup$
Suppose I have a set of elements $$ S = lbrace A,B,B,D,E rbrace$$
Then the number of distinct permutations of size 5 is
$$frac{5!}{2!} = 60$$
This is because $S$ contains a ‘copy’ of B.
But by definition, in a set of elements, each element is distinct. However, now I’m essentially saying “wait actually B and B aren’t distinct.”
So why isn’t the set $$ S = lbrace A,B,D,E rbrace$$ if the B elements are not distinct? Notice the permutations of this set are very different
$$4! = 24$$
Are there two different kinds of being distinct here?
combinatorics
asked Dec 17 '18 at 22:29
Stan ShunpikeStan Shunpike
1,83111439
$endgroup$
3
$begingroup$
You should think of $S$ as a multiset, not a set; one in which multiple copies do matter. Alternatively, put a yellow sticker on one of the $B$s and a green sticker on the other one so that they are now distinct. Count the number of permutations; then remove the stickets. Figure out how many ways the final permutation could have occurred to get the correct divisor.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 22:43
2
$begingroup$
@ArturoMagidin Can you add this as an answer? I think this is what I am looking for. Perhaps expand a bit if you can.
$endgroup$
– Stan Shunpike
Dec 17 '18 at 22:49
add a comment |
$begingroup$
Suppose I have a set of elements $$ S = lbrace A,B,B,D,E rbrace$$
Then the number of distinct permutations of size 5 is
$$frac{5!}{2!} = 60$$
This is because $S$ contains a ‘copy’ of B.
But by definition, in a set of elements, each element is distinct. However, now I’m essentially saying “wait actually B and B aren’t distinct.”
So why isn’t the set $$ S = lbrace A,B,D,E rbrace$$ if the B elements are not distinct? Notice the permutations of this set are very different
$$4! = 24$$
Are there two different kinds of being distinct here?
combinatorics
asked Dec 17 '18 at 22:29
Stan ShunpikeStan Shunpike
1,83111439
$endgroup$
Suppose I have a set of elements $$ S = lbrace A,B,B,D,E rbrace$$
Then the number of distinct permutations of size 5 is
$$frac{5!}{2!} = 60$$
This is because $S$ contains a ‘copy’ of B.
But by definition, in a set of elements, each element is distinct. However, now I’m essentially saying “wait actually B and B aren’t distinct.”
So why isn’t the set $$ S = lbrace A,B,D,E rbrace$$ if the B elements are not distinct? Notice the permutations of this set are very different
$$4! = 24$$
Are there two different kinds of being distinct here?
combinatorics
combinatorics
asked Dec 17 '18 at 22:29
Stan ShunpikeStan Shunpike
1,83111439
asked Dec 17 '18 at 22:29
Stan ShunpikeStan Shunpike
1,83111439
asked Dec 17 '18 at 22:29
Stan ShunpikeStan Shunpike
1,83111439
asked Dec 17 '18 at 22:29
Stan ShunpikeStan Shunpike
1,83111439
asked Dec 17 '18 at 22:29
Stan ShunpikeStan Shunpike
1,83111439
1,83111439
3
$begingroup$
You should think of $S$ as a multiset, not a set; one in which multiple copies do matter. Alternatively, put a yellow sticker on one of the $B$s and a green sticker on the other one so that they are now distinct. Count the number of permutations; then remove the stickets. Figure out how many ways the final permutation could have occurred to get the correct divisor.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 22:43
2
$begingroup$
@ArturoMagidin Can you add this as an answer? I think this is what I am looking for. Perhaps expand a bit if you can.
$endgroup$
– Stan Shunpike
Dec 17 '18 at 22:49
add a comment |
3
$begingroup$
You should think of $S$ as a multiset, not a set; one in which multiple copies do matter. Alternatively, put a yellow sticker on one of the $B$s and a green sticker on the other one so that they are now distinct. Count the number of permutations; then remove the stickets. Figure out how many ways the final permutation could have occurred to get the correct divisor.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 22:43
2
$begingroup$
@ArturoMagidin Can you add this as an answer? I think this is what I am looking for. Perhaps expand a bit if you can.
$endgroup$
– Stan Shunpike
Dec 17 '18 at 22:49
3
3
$begingroup$
You should think of $S$ as a multiset, not a set; one in which multiple copies do matter. Alternatively, put a yellow sticker on one of the $B$s and a green sticker on the other one so that they are now distinct. Count the number of permutations; then remove the stickets. Figure out how many ways the final permutation could have occurred to get the correct divisor.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 22:43
$begingroup$
You should think of $S$ as a multiset, not a set; one in which multiple copies do matter. Alternatively, put a yellow sticker on one of the $B$s and a green sticker on the other one so that they are now distinct. Count the number of permutations; then remove the stickets. Figure out how many ways the final permutation could have occurred to get the correct divisor.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 22:43
2
2
$begingroup$
@ArturoMagidin Can you add this as an answer? I think this is what I am looking for. Perhaps expand a bit if you can.
$endgroup$
– Stan Shunpike
Dec 17 '18 at 22:49
$begingroup$
@ArturoMagidin Can you add this as an answer? I think this is what I am looking for. Perhaps expand a bit if you can.
$endgroup$
– Stan Shunpike
Dec 17 '18 at 22:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You should see the counts as the number of outputs of two different experiments:
First we have a bag of 5 letter tiles (a la Scrabble say) with $A,B,B,D,E$ tiles.
We draw 5 tiles from the bag and put them in a row in order we draw them. We have 5 draws, so $5!$ many permutations of the tiles. But in reality we cannot (in the end) distinguish the two B's (We could have marked them and then we'd have $5!$ different results): so the number of end results (sequences of 5 letters) is halved, because for every permutation $ldots B_1ldots B_2ldots$ have a seocnd equivalent one $ldots B_2ldots B_1ldots$.
The second experiment we have a bag with four tiles $A,B,D,E$ and counts the number of possible permutations from a draw of $4$ letters put in a row, so there we count sequences of $4$ letters, not of $5$. Hence the difference in counts.
answered Dec 17 '18 at 22:40
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
add a comment |
$begingroup$
take a simpler case ${A,B,B}$ vs ${A,B}$. First one will have $ABB,BBA,BAB$, however the second one only $AB,BA$. Note that when $BB$ are next to each other it can be mapped to the single instance case one-to-one, but not when there are separated.
answered Dec 17 '18 at 22:42
karakfakarakfa
2,025811
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
You should see the counts as the number of outputs of two different experiments:
First we have a bag of 5 letter tiles (a la Scrabble say) with $A,B,B,D,E$ tiles.
We draw 5 tiles from the bag and put them in a row in order we draw them. We have 5 draws, so $5!$ many permutations of the tiles. But in reality we cannot (in the end) distinguish the two B's (We could have marked them and then we'd have $5!$ different results): so the number of end results (sequences of 5 letters) is halved, because for every permutation $ldots B_1ldots B_2ldots$ have a seocnd equivalent one $ldots B_2ldots B_1ldots$.
The second experiment we have a bag with four tiles $A,B,D,E$ and counts the number of possible permutations from a draw of $4$ letters put in a row, so there we count sequences of $4$ letters, not of $5$. Hence the difference in counts.
answered Dec 17 '18 at 22:40
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
add a comment |
$begingroup$
You should see the counts as the number of outputs of two different experiments:
First we have a bag of 5 letter tiles (a la Scrabble say) with $A,B,B,D,E$ tiles.
We draw 5 tiles from the bag and put them in a row in order we draw them. We have 5 draws, so $5!$ many permutations of the tiles. But in reality we cannot (in the end) distinguish the two B's (We could have marked them and then we'd have $5!$ different results): so the number of end results (sequences of 5 letters) is halved, because for every permutation $ldots B_1ldots B_2ldots$ have a seocnd equivalent one $ldots B_2ldots B_1ldots$.
The second experiment we have a bag with four tiles $A,B,D,E$ and counts the number of possible permutations from a draw of $4$ letters put in a row, so there we count sequences of $4$ letters, not of $5$. Hence the difference in counts.
answered Dec 17 '18 at 22:40
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
add a comment |
$begingroup$
You should see the counts as the number of outputs of two different experiments:
First we have a bag of 5 letter tiles (a la Scrabble say) with $A,B,B,D,E$ tiles.
We draw 5 tiles from the bag and put them in a row in order we draw them. We have 5 draws, so $5!$ many permutations of the tiles. But in reality we cannot (in the end) distinguish the two B's (We could have marked them and then we'd have $5!$ different results): so the number of end results (sequences of 5 letters) is halved, because for every permutation $ldots B_1ldots B_2ldots$ have a seocnd equivalent one $ldots B_2ldots B_1ldots$.
The second experiment we have a bag with four tiles $A,B,D,E$ and counts the number of possible permutations from a draw of $4$ letters put in a row, so there we count sequences of $4$ letters, not of $5$. Hence the difference in counts.
answered Dec 17 '18 at 22:40
Henno BrandsmaHenno Brandsma
113k348123
$endgroup$
You should see the counts as the number of outputs of two different experiments:
First we have a bag of 5 letter tiles (a la Scrabble say) with $A,B,B,D,E$ tiles.
We draw 5 tiles from the bag and put them in a row in order we draw them. We have 5 draws, so $5!$ many permutations of the tiles. But in reality we cannot (in the end) distinguish the two B's (We could have marked them and then we'd have $5!$ different results): so the number of end results (sequences of 5 letters) is halved, because for every permutation $ldots B_1ldots B_2ldots$ have a seocnd equivalent one $ldots B_2ldots B_1ldots$.
The second experiment we have a bag with four tiles $A,B,D,E$ and counts the number of possible permutations from a draw of $4$ letters put in a row, so there we count sequences of $4$ letters, not of $5$. Hence the difference in counts.
answered Dec 17 '18 at 22:40
Henno BrandsmaHenno Brandsma
113k348123
answered Dec 17 '18 at 22:40
Henno BrandsmaHenno Brandsma
113k348123
answered Dec 17 '18 at 22:40
Henno BrandsmaHenno Brandsma
113k348123
answered Dec 17 '18 at 22:40
Henno BrandsmaHenno Brandsma
113k348123
113k348123
add a comment |
add a comment |
$begingroup$
take a simpler case ${A,B,B}$ vs ${A,B}$. First one will have $ABB,BBA,BAB$, however the second one only $AB,BA$. Note that when $BB$ are next to each other it can be mapped to the single instance case one-to-one, but not when there are separated.
answered Dec 17 '18 at 22:42
karakfakarakfa
2,025811
$endgroup$
add a comment |
$begingroup$
take a simpler case ${A,B,B}$ vs ${A,B}$. First one will have $ABB,BBA,BAB$, however the second one only $AB,BA$. Note that when $BB$ are next to each other it can be mapped to the single instance case one-to-one, but not when there are separated.
answered Dec 17 '18 at 22:42
karakfakarakfa
2,025811
$endgroup$
add a comment |
$begingroup$
take a simpler case ${A,B,B}$ vs ${A,B}$. First one will have $ABB,BBA,BAB$, however the second one only $AB,BA$. Note that when $BB$ are next to each other it can be mapped to the single instance case one-to-one, but not when there are separated.
answered Dec 17 '18 at 22:42
karakfakarakfa
2,025811
$endgroup$
take a simpler case ${A,B,B}$ vs ${A,B}$. First one will have $ABB,BBA,BAB$, however the second one only $AB,BA$. Note that when $BB$ are next to each other it can be mapped to the single instance case one-to-one, but not when there are separated.
answered Dec 17 '18 at 22:42
karakfakarakfa
2,025811
answered Dec 17 '18 at 22:42
karakfakarakfa
2,025811
answered Dec 17 '18 at 22:42
karakfakarakfa
2,025811
answered Dec 17 '18 at 22:42
karakfakarakfa
2,025811
2,025811
add a comment |
add a comment |
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$begingroup$
You should think of $S$ as a multiset, not a set; one in which multiple copies do matter. Alternatively, put a yellow sticker on one of the $B$s and a green sticker on the other one so that they are now distinct. Count the number of permutations; then remove the stickets. Figure out how many ways the final permutation could have occurred to get the correct divisor.
$endgroup$
– Arturo Magidin
Dec 17 '18 at 22:43
2
$begingroup$
@ArturoMagidin Can you add this as an answer? I think this is what I am looking for. Perhaps expand a bit if you can.
$endgroup$
– Stan Shunpike
Dec 17 '18 at 22:49