For permutations with “copies”, in what sense are the elements of the set distinct?












2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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


















2












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















2












2








2





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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
















  • 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












2 Answers
2






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2












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    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.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      2












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          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.






          share|cite|improve this answer









          $endgroup$



          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.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 17 '18 at 22:40









          Henno BrandsmaHenno Brandsma

          113k348123




          113k348123























              1












              $begingroup$

              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.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                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.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  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.






                  share|cite|improve this answer









                  $endgroup$



                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 17 '18 at 22:42









                  karakfakarakfa

                  2,025811




                  2,025811






























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