Number of ways to arrange the letters BANANAS












0














I was in class, and the teacher told us, that the number of ways to re-arrange the letters in BANANAS is $frac{7!}{2!3!)}$. I understand where the $2, 3$ and $7$ are coming from (the number of $N$'s, the number of $A$'s, and the total number of letters respectively), and why we are using factorials (the number of ways to arrange $n$ things is $n!$), but I still feel like I am missing something crucial.



I suppose, why is it that we are dividing by $2!$ and $3!$?
Why is it that $2!$ and $3!$ are being multiplied together?






combinatorics permutations







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  • We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
    – Love Invariants
    Dec 5 '18 at 16:24








  • 1




    Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
    – Robert Z
    Dec 5 '18 at 16:25
















0














I was in class, and the teacher told us, that the number of ways to re-arrange the letters in BANANAS is $frac{7!}{2!3!)}$. I understand where the $2, 3$ and $7$ are coming from (the number of $N$'s, the number of $A$'s, and the total number of letters respectively), and why we are using factorials (the number of ways to arrange $n$ things is $n!$), but I still feel like I am missing something crucial.



I suppose, why is it that we are dividing by $2!$ and $3!$?
Why is it that $2!$ and $3!$ are being multiplied together?






combinatorics permutations







share|cite|improve this question
























  • We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
    – Love Invariants
    Dec 5 '18 at 16:24








  • 1




    Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
    – Robert Z
    Dec 5 '18 at 16:25














0












0








0







I was in class, and the teacher told us, that the number of ways to re-arrange the letters in BANANAS is $frac{7!}{2!3!)}$. I understand where the $2, 3$ and $7$ are coming from (the number of $N$'s, the number of $A$'s, and the total number of letters respectively), and why we are using factorials (the number of ways to arrange $n$ things is $n!$), but I still feel like I am missing something crucial.



I suppose, why is it that we are dividing by $2!$ and $3!$?
Why is it that $2!$ and $3!$ are being multiplied together?






combinatorics permutations







share|cite|improve this question















I was in class, and the teacher told us, that the number of ways to re-arrange the letters in BANANAS is $frac{7!}{2!3!)}$. I understand where the $2, 3$ and $7$ are coming from (the number of $N$'s, the number of $A$'s, and the total number of letters respectively), and why we are using factorials (the number of ways to arrange $n$ things is $n!$), but I still feel like I am missing something crucial.



I suppose, why is it that we are dividing by $2!$ and $3!$?
Why is it that $2!$ and $3!$ are being multiplied together?






combinatorics permutations




combinatorics permutations






share|cite|improve this question















share|cite|improve this question













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edited Dec 5 '18 at 16:57









user3482749

3,196414




3,196414










asked Dec 5 '18 at 16:20









nessness

375




375












  • We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
    – Love Invariants
    Dec 5 '18 at 16:24








  • 1




    Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
    – Robert Z
    Dec 5 '18 at 16:25


















  • We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
    – Love Invariants
    Dec 5 '18 at 16:24








  • 1




    Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
    – Robert Z
    Dec 5 '18 at 16:25
















We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
– Love Invariants
Dec 5 '18 at 16:24






We are dividing by 2! and 3! simply because all 3 A's are alike letters and 2 N's are alike. They can be shuffled in 3! and 2! ways respectively.
– Love Invariants
Dec 5 '18 at 16:24






1




1




Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
– Robert Z
Dec 5 '18 at 16:25




Try with a smaller case like $ANANA$. Write down all the arrangements. Their number should be $5!/2!/3!=10$
– Robert Z
Dec 5 '18 at 16:25










2 Answers
2






active

oldest

votes


















6














Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.



Then the number of arrangements is $7!$. Imagine that you write down all of them.



Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.






share|cite|improve this answer































    2














    Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
    $$
    BA_1N_1A_2N_2A_3S
    $$
    Then there truly are $7!$ ways to rearrange this word.



    However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
    $$
    BA_1N_1A_2N_2A_3S\
    BA_1N_2A_2N_1A_3S
    $$
    will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.



    Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
    $$
    BA_1NA_2NA_3S\
    BA_1NA_3NA_2S\
    BA_2NA_1NA_3S\
    BA_2NA_3NA_1S\
    BA_3NA_1NA_2S\
    BA_3NA_2NA_1S
    $$
    will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      6














      Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.



      Then the number of arrangements is $7!$. Imagine that you write down all of them.



      Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.






      share|cite|improve this answer




























        6














        Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.



        Then the number of arrangements is $7!$. Imagine that you write down all of them.



        Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.






        share|cite|improve this answer


























          6












          6








          6






          Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.



          Then the number of arrangements is $7!$. Imagine that you write down all of them.



          Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.






          share|cite|improve this answer














          Assume that all the letter are different. For example, imagine that each $A$ and $N$ is colored differently.



          Then the number of arrangements is $7!$. Imagine that you write down all of them.



          Now return all colored letters to black, and note that there are many 'words' that are counted multiple times. Namely, every word is counted $2!3!$ times, exactly the number of ways to rearrange the $A$'s and $N$'s in it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 5 '18 at 17:17









          AryanSonwatikar

          31312




          31312










          answered Dec 5 '18 at 16:24









          ajotatxeajotatxe

          53.6k23890




          53.6k23890























              2














              Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
              $$
              BA_1N_1A_2N_2A_3S
              $$
              Then there truly are $7!$ ways to rearrange this word.



              However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
              $$
              BA_1N_1A_2N_2A_3S\
              BA_1N_2A_2N_1A_3S
              $$
              will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.



              Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
              $$
              BA_1NA_2NA_3S\
              BA_1NA_3NA_2S\
              BA_2NA_1NA_3S\
              BA_2NA_3NA_1S\
              BA_3NA_1NA_2S\
              BA_3NA_2NA_1S
              $$
              will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.






              share|cite|improve this answer


























                2














                Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
                $$
                BA_1N_1A_2N_2A_3S
                $$
                Then there truly are $7!$ ways to rearrange this word.



                However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
                $$
                BA_1N_1A_2N_2A_3S\
                BA_1N_2A_2N_1A_3S
                $$
                will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.



                Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
                $$
                BA_1NA_2NA_3S\
                BA_1NA_3NA_2S\
                BA_2NA_1NA_3S\
                BA_2NA_3NA_1S\
                BA_3NA_1NA_2S\
                BA_3NA_2NA_1S
                $$
                will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.






                share|cite|improve this answer
























                  2












                  2








                  2






                  Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
                  $$
                  BA_1N_1A_2N_2A_3S
                  $$
                  Then there truly are $7!$ ways to rearrange this word.



                  However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
                  $$
                  BA_1N_1A_2N_2A_3S\
                  BA_1N_2A_2N_1A_3S
                  $$
                  will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.



                  Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
                  $$
                  BA_1NA_2NA_3S\
                  BA_1NA_3NA_2S\
                  BA_2NA_1NA_3S\
                  BA_2NA_3NA_1S\
                  BA_3NA_1NA_2S\
                  BA_3NA_2NA_1S
                  $$
                  will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.






                  share|cite|improve this answer












                  Consider making the $A$'s nd the $N$'s different by, say, putting a label on them, like so:
                  $$
                  BA_1N_1A_2N_2A_3S
                  $$
                  Then there truly are $7!$ ways to rearrange this word.



                  However, once we remove the labels, some of these $7!$ arrangements collapse into the same arrangement. Let's remove the labels on the $N$'s first. In that case, for instance,
                  $$
                  BA_1N_1A_2N_2A_3S\
                  BA_1N_2A_2N_1A_3S
                  $$
                  will become the same. In fact, all the $7!$ arrangements group together two by two in pairs which become the same after removing the labels from the $N$'s. The number of distinct words after removing the $N$ labels is therefore $frac{7!}{2}$.



                  Now let's look at these $frac{7!}{2}$ words, and remove the labels from the $A$'s as well. Again, many different arrangements will collapse into the same arrangement. For instance,
                  $$
                  BA_1NA_2NA_3S\
                  BA_1NA_3NA_2S\
                  BA_2NA_1NA_3S\
                  BA_2NA_3NA_1S\
                  BA_3NA_1NA_2S\
                  BA_3NA_2NA_1S
                  $$
                  will all become the same word after removing the labels. In fact, all the $frac{7!}{2}$ different arrangements will go together in groups of $6$ where all the words in one group become equal after removing the label. There are $frac{7!}{2cdot 6}$ such groups, so that's the number of distinguishable words after removing all the labels from both the $N$'s and the $A$'s.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 5 '18 at 16:30









                  ArthurArthur

                  111k7107189




                  111k7107189






























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