How many words can be made using the letters from TALLAHASSEE?











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This question has already been answered (how is it combination prob, Trouble in finding number of permutations, Trouble in finding number of permutations), I first seen it in my discrete maths book, and I understood it.



But, since this is a really difficult topic (combinatorics) for me, I was watching this video and the same problem came up but they changed it a little bit.




Given the word TALLAHASSEE, how many arrangaments can be made with no consecutive vowels?




He kinda uses the same approach as the book, but the last thing he does is to order the case of $AAAEE$, which is obviously $frac{5!}{3!2!}$. But this doesn't make sense to me at all, the problem says that you can't have any conscutive vowels, so he's counting them by doing that?



For me, the correct answer would be $frac{6!}{2!2!}{7choose 5}$, where $frac{6!}{2!2!}$ is the number of arrangements with no $A's$ and no $E's$ and ${7choose 5}$ are the positions where the vowels could be in between the rest of the letters.



Can someone explain this to me a little bit?










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  • 2




    In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).
    – rogerl
    Dec 4 at 16:12










  • @rogerl but when you do $frac{5!}{3!2!}$ and then you multiply it by the rest of the things, aren't you counting the cases where for example, two $A's$ could be together? Or $AE's$, etc?
    – parishilton
    Dec 4 at 16:15

















up vote
0
down vote

favorite
1












This question has already been answered (how is it combination prob, Trouble in finding number of permutations, Trouble in finding number of permutations), I first seen it in my discrete maths book, and I understood it.



But, since this is a really difficult topic (combinatorics) for me, I was watching this video and the same problem came up but they changed it a little bit.




Given the word TALLAHASSEE, how many arrangaments can be made with no consecutive vowels?




He kinda uses the same approach as the book, but the last thing he does is to order the case of $AAAEE$, which is obviously $frac{5!}{3!2!}$. But this doesn't make sense to me at all, the problem says that you can't have any conscutive vowels, so he's counting them by doing that?



For me, the correct answer would be $frac{6!}{2!2!}{7choose 5}$, where $frac{6!}{2!2!}$ is the number of arrangements with no $A's$ and no $E's$ and ${7choose 5}$ are the positions where the vowels could be in between the rest of the letters.



Can someone explain this to me a little bit?










share|cite|improve this question


















  • 2




    In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).
    – rogerl
    Dec 4 at 16:12










  • @rogerl but when you do $frac{5!}{3!2!}$ and then you multiply it by the rest of the things, aren't you counting the cases where for example, two $A's$ could be together? Or $AE's$, etc?
    – parishilton
    Dec 4 at 16:15















up vote
0
down vote

favorite
1









up vote
0
down vote

favorite
1






1





This question has already been answered (how is it combination prob, Trouble in finding number of permutations, Trouble in finding number of permutations), I first seen it in my discrete maths book, and I understood it.



But, since this is a really difficult topic (combinatorics) for me, I was watching this video and the same problem came up but they changed it a little bit.




Given the word TALLAHASSEE, how many arrangaments can be made with no consecutive vowels?




He kinda uses the same approach as the book, but the last thing he does is to order the case of $AAAEE$, which is obviously $frac{5!}{3!2!}$. But this doesn't make sense to me at all, the problem says that you can't have any conscutive vowels, so he's counting them by doing that?



For me, the correct answer would be $frac{6!}{2!2!}{7choose 5}$, where $frac{6!}{2!2!}$ is the number of arrangements with no $A's$ and no $E's$ and ${7choose 5}$ are the positions where the vowels could be in between the rest of the letters.



Can someone explain this to me a little bit?










share|cite|improve this question













This question has already been answered (how is it combination prob, Trouble in finding number of permutations, Trouble in finding number of permutations), I first seen it in my discrete maths book, and I understood it.



But, since this is a really difficult topic (combinatorics) for me, I was watching this video and the same problem came up but they changed it a little bit.




Given the word TALLAHASSEE, how many arrangaments can be made with no consecutive vowels?




He kinda uses the same approach as the book, but the last thing he does is to order the case of $AAAEE$, which is obviously $frac{5!}{3!2!}$. But this doesn't make sense to me at all, the problem says that you can't have any conscutive vowels, so he's counting them by doing that?



For me, the correct answer would be $frac{6!}{2!2!}{7choose 5}$, where $frac{6!}{2!2!}$ is the number of arrangements with no $A's$ and no $E's$ and ${7choose 5}$ are the positions where the vowels could be in between the rest of the letters.



Can someone explain this to me a little bit?







combinatorics discrete-mathematics combinations






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asked Dec 4 at 16:02









parishilton

15910




15910








  • 2




    In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).
    – rogerl
    Dec 4 at 16:12










  • @rogerl but when you do $frac{5!}{3!2!}$ and then you multiply it by the rest of the things, aren't you counting the cases where for example, two $A's$ could be together? Or $AE's$, etc?
    – parishilton
    Dec 4 at 16:15
















  • 2




    In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).
    – rogerl
    Dec 4 at 16:12










  • @rogerl but when you do $frac{5!}{3!2!}$ and then you multiply it by the rest of the things, aren't you counting the cases where for example, two $A's$ could be together? Or $AE's$, etc?
    – parishilton
    Dec 4 at 16:15










2




2




In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).
– rogerl
Dec 4 at 16:12




In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).
– rogerl
Dec 4 at 16:12












@rogerl but when you do $frac{5!}{3!2!}$ and then you multiply it by the rest of the things, aren't you counting the cases where for example, two $A's$ could be together? Or $AE's$, etc?
– parishilton
Dec 4 at 16:15






@rogerl but when you do $frac{5!}{3!2!}$ and then you multiply it by the rest of the things, aren't you counting the cases where for example, two $A's$ could be together? Or $AE's$, etc?
– parishilton
Dec 4 at 16:15












2 Answers
2






active

oldest

votes

















up vote
1
down vote



accepted










In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).



To think about it a different way, suppose all the vowels were some symbol, say $*$. Then the number of arrangements with no $*$'s together would be, as you computed, $frac{6!}{2!2!}binom{7}{5}$. Now you have five $*$'s that you need to replace with three $A$s and two $E$s. In how many ways can you do this? Well, you just need to choose the locations for the $A$s, which can be done in $binom{5}{3}$ ways.






share|cite|improve this answer





















  • Oooo I got it now! Thank you soooo much!!!
    – parishilton
    Dec 4 at 16:25


















up vote
0
down vote













You should break it into three separate steps like it was done in the video.



1) Number of ways to arrange $TLLHSS$:
$$frac{6!}{2!2!}$$
2) Number of ways to arrange five vowels into $7$ positions in $__ T__ L__ L__ H__ S__ S__ $ one by one:
$${7choose 5}$$
Note that here we are only interested in whether we should place a vowel in a specific position or not. In other words, we are just marking the positions for vowels (not specifying them yet). For example, one arrangement is
$$__ T V_1 L V_2 L V_3 H __ S V_4 S V_5,$$
where $V_i$ is a vowel.



3) Number of ways to arrange $AAAEE$:
$$frac{5!}{3!2!}$$
Note that for any five positions we want to place $5$ vowels in 2) there are $frac{5!}{3!2!}$ arrangements to place $AAAEE$. It implies
$$__ Tcolor{red}ALcolor{blue}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and $$__ Tcolor{blue}ALcolor{red}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and all other arrangements of $AAAEE$ in this order and in the same positions are the same arrangement.



That is it.



I have watched many videos of TheTrevTutor and must admit it is a great resource!






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    2 Answers
    2






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    2 Answers
    2






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    up vote
    1
    down vote



    accepted










    In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).



    To think about it a different way, suppose all the vowels were some symbol, say $*$. Then the number of arrangements with no $*$'s together would be, as you computed, $frac{6!}{2!2!}binom{7}{5}$. Now you have five $*$'s that you need to replace with three $A$s and two $E$s. In how many ways can you do this? Well, you just need to choose the locations for the $A$s, which can be done in $binom{5}{3}$ ways.






    share|cite|improve this answer





















    • Oooo I got it now! Thank you soooo much!!!
      – parishilton
      Dec 4 at 16:25















    up vote
    1
    down vote



    accepted










    In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).



    To think about it a different way, suppose all the vowels were some symbol, say $*$. Then the number of arrangements with no $*$'s together would be, as you computed, $frac{6!}{2!2!}binom{7}{5}$. Now you have five $*$'s that you need to replace with three $A$s and two $E$s. In how many ways can you do this? Well, you just need to choose the locations for the $A$s, which can be done in $binom{5}{3}$ ways.






    share|cite|improve this answer





















    • Oooo I got it now! Thank you soooo much!!!
      – parishilton
      Dec 4 at 16:25













    up vote
    1
    down vote



    accepted







    up vote
    1
    down vote



    accepted






    In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).



    To think about it a different way, suppose all the vowels were some symbol, say $*$. Then the number of arrangements with no $*$'s together would be, as you computed, $frac{6!}{2!2!}binom{7}{5}$. Now you have five $*$'s that you need to replace with three $A$s and two $E$s. In how many ways can you do this? Well, you just need to choose the locations for the $A$s, which can be done in $binom{5}{3}$ ways.






    share|cite|improve this answer












    In the original post you cite, all arrangements of the A's are equivalent (that is, all the symbols are the same). Here, though, putting AAAEE into five distinct holes gives a different word that putting EEAAA into those same holes (in order).



    To think about it a different way, suppose all the vowels were some symbol, say $*$. Then the number of arrangements with no $*$'s together would be, as you computed, $frac{6!}{2!2!}binom{7}{5}$. Now you have five $*$'s that you need to replace with three $A$s and two $E$s. In how many ways can you do this? Well, you just need to choose the locations for the $A$s, which can be done in $binom{5}{3}$ ways.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 4 at 16:21









    rogerl

    17.3k22746




    17.3k22746












    • Oooo I got it now! Thank you soooo much!!!
      – parishilton
      Dec 4 at 16:25


















    • Oooo I got it now! Thank you soooo much!!!
      – parishilton
      Dec 4 at 16:25
















    Oooo I got it now! Thank you soooo much!!!
    – parishilton
    Dec 4 at 16:25




    Oooo I got it now! Thank you soooo much!!!
    – parishilton
    Dec 4 at 16:25










    up vote
    0
    down vote













    You should break it into three separate steps like it was done in the video.



    1) Number of ways to arrange $TLLHSS$:
    $$frac{6!}{2!2!}$$
    2) Number of ways to arrange five vowels into $7$ positions in $__ T__ L__ L__ H__ S__ S__ $ one by one:
    $${7choose 5}$$
    Note that here we are only interested in whether we should place a vowel in a specific position or not. In other words, we are just marking the positions for vowels (not specifying them yet). For example, one arrangement is
    $$__ T V_1 L V_2 L V_3 H __ S V_4 S V_5,$$
    where $V_i$ is a vowel.



    3) Number of ways to arrange $AAAEE$:
    $$frac{5!}{3!2!}$$
    Note that for any five positions we want to place $5$ vowels in 2) there are $frac{5!}{3!2!}$ arrangements to place $AAAEE$. It implies
    $$__ Tcolor{red}ALcolor{blue}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and $$__ Tcolor{blue}ALcolor{red}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and all other arrangements of $AAAEE$ in this order and in the same positions are the same arrangement.



    That is it.



    I have watched many videos of TheTrevTutor and must admit it is a great resource!






    share|cite|improve this answer

























      up vote
      0
      down vote













      You should break it into three separate steps like it was done in the video.



      1) Number of ways to arrange $TLLHSS$:
      $$frac{6!}{2!2!}$$
      2) Number of ways to arrange five vowels into $7$ positions in $__ T__ L__ L__ H__ S__ S__ $ one by one:
      $${7choose 5}$$
      Note that here we are only interested in whether we should place a vowel in a specific position or not. In other words, we are just marking the positions for vowels (not specifying them yet). For example, one arrangement is
      $$__ T V_1 L V_2 L V_3 H __ S V_4 S V_5,$$
      where $V_i$ is a vowel.



      3) Number of ways to arrange $AAAEE$:
      $$frac{5!}{3!2!}$$
      Note that for any five positions we want to place $5$ vowels in 2) there are $frac{5!}{3!2!}$ arrangements to place $AAAEE$. It implies
      $$__ Tcolor{red}ALcolor{blue}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and $$__ Tcolor{blue}ALcolor{red}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and all other arrangements of $AAAEE$ in this order and in the same positions are the same arrangement.



      That is it.



      I have watched many videos of TheTrevTutor and must admit it is a great resource!






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        You should break it into three separate steps like it was done in the video.



        1) Number of ways to arrange $TLLHSS$:
        $$frac{6!}{2!2!}$$
        2) Number of ways to arrange five vowels into $7$ positions in $__ T__ L__ L__ H__ S__ S__ $ one by one:
        $${7choose 5}$$
        Note that here we are only interested in whether we should place a vowel in a specific position or not. In other words, we are just marking the positions for vowels (not specifying them yet). For example, one arrangement is
        $$__ T V_1 L V_2 L V_3 H __ S V_4 S V_5,$$
        where $V_i$ is a vowel.



        3) Number of ways to arrange $AAAEE$:
        $$frac{5!}{3!2!}$$
        Note that for any five positions we want to place $5$ vowels in 2) there are $frac{5!}{3!2!}$ arrangements to place $AAAEE$. It implies
        $$__ Tcolor{red}ALcolor{blue}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and $$__ Tcolor{blue}ALcolor{red}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and all other arrangements of $AAAEE$ in this order and in the same positions are the same arrangement.



        That is it.



        I have watched many videos of TheTrevTutor and must admit it is a great resource!






        share|cite|improve this answer












        You should break it into three separate steps like it was done in the video.



        1) Number of ways to arrange $TLLHSS$:
        $$frac{6!}{2!2!}$$
        2) Number of ways to arrange five vowels into $7$ positions in $__ T__ L__ L__ H__ S__ S__ $ one by one:
        $${7choose 5}$$
        Note that here we are only interested in whether we should place a vowel in a specific position or not. In other words, we are just marking the positions for vowels (not specifying them yet). For example, one arrangement is
        $$__ T V_1 L V_2 L V_3 H __ S V_4 S V_5,$$
        where $V_i$ is a vowel.



        3) Number of ways to arrange $AAAEE$:
        $$frac{5!}{3!2!}$$
        Note that for any five positions we want to place $5$ vowels in 2) there are $frac{5!}{3!2!}$ arrangements to place $AAAEE$. It implies
        $$__ Tcolor{red}ALcolor{blue}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and $$__ Tcolor{blue}ALcolor{red}ALcolor{green}AH __ Scolor{red}EScolor{blue}E$$ and all other arrangements of $AAAEE$ in this order and in the same positions are the same arrangement.



        That is it.



        I have watched many videos of TheTrevTutor and must admit it is a great resource!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 4 at 18:03









        farruhota

        18.5k2736




        18.5k2736






























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