Counting question on bit strings - problem with using cases











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How many bit strings of length 10 either begin with three 0s or end with two 0s?




I solved this question using cases but I do not seem to be getting the answer of $352$.



My attempt:
Consider two cases:




  • Case 1: The string begins with three $0$s and does not end with two $0$s. There is only $1$ way to choose the first three bits, $2^5$ ways for the middle bits, and $3$ ways for the last two bits ($4$ ways to construct a string of two bits, minus $1$ way to make three $0$s). There are $2^5 cdot 3$ ways to construct strings of this type.

  • Case 2: The string does not begin with three $0$s but ends with two $0$s. There are $2^3 - 1 = 7$ ways to choose the first three bits without three $0$s, $2^5$ ways for the middle bits, and $1$ way for the last two bits. There are $7 cdot 2^5$ ways to construct strings of this type.


By the rule of sum, there are $2^5 cdot 3 + 2^5 cdot 7 = 320$ ways to construct bit strings of length 10 either begin with three $0$s or end with two $0$s.










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    up vote
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    down vote

    favorite













    How many bit strings of length 10 either begin with three 0s or end with two 0s?




    I solved this question using cases but I do not seem to be getting the answer of $352$.



    My attempt:
    Consider two cases:




    • Case 1: The string begins with three $0$s and does not end with two $0$s. There is only $1$ way to choose the first three bits, $2^5$ ways for the middle bits, and $3$ ways for the last two bits ($4$ ways to construct a string of two bits, minus $1$ way to make three $0$s). There are $2^5 cdot 3$ ways to construct strings of this type.

    • Case 2: The string does not begin with three $0$s but ends with two $0$s. There are $2^3 - 1 = 7$ ways to choose the first three bits without three $0$s, $2^5$ ways for the middle bits, and $1$ way for the last two bits. There are $7 cdot 2^5$ ways to construct strings of this type.


    By the rule of sum, there are $2^5 cdot 3 + 2^5 cdot 7 = 320$ ways to construct bit strings of length 10 either begin with three $0$s or end with two $0$s.










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      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite












      How many bit strings of length 10 either begin with three 0s or end with two 0s?




      I solved this question using cases but I do not seem to be getting the answer of $352$.



      My attempt:
      Consider two cases:




      • Case 1: The string begins with three $0$s and does not end with two $0$s. There is only $1$ way to choose the first three bits, $2^5$ ways for the middle bits, and $3$ ways for the last two bits ($4$ ways to construct a string of two bits, minus $1$ way to make three $0$s). There are $2^5 cdot 3$ ways to construct strings of this type.

      • Case 2: The string does not begin with three $0$s but ends with two $0$s. There are $2^3 - 1 = 7$ ways to choose the first three bits without three $0$s, $2^5$ ways for the middle bits, and $1$ way for the last two bits. There are $7 cdot 2^5$ ways to construct strings of this type.


      By the rule of sum, there are $2^5 cdot 3 + 2^5 cdot 7 = 320$ ways to construct bit strings of length 10 either begin with three $0$s or end with two $0$s.










      share|cite|improve this question














      How many bit strings of length 10 either begin with three 0s or end with two 0s?




      I solved this question using cases but I do not seem to be getting the answer of $352$.



      My attempt:
      Consider two cases:




      • Case 1: The string begins with three $0$s and does not end with two $0$s. There is only $1$ way to choose the first three bits, $2^5$ ways for the middle bits, and $3$ ways for the last two bits ($4$ ways to construct a string of two bits, minus $1$ way to make three $0$s). There are $2^5 cdot 3$ ways to construct strings of this type.

      • Case 2: The string does not begin with three $0$s but ends with two $0$s. There are $2^3 - 1 = 7$ ways to choose the first three bits without three $0$s, $2^5$ ways for the middle bits, and $1$ way for the last two bits. There are $7 cdot 2^5$ ways to construct strings of this type.


      By the rule of sum, there are $2^5 cdot 3 + 2^5 cdot 7 = 320$ ways to construct bit strings of length 10 either begin with three $0$s or end with two $0$s.







      combinatorics discrete-mathematics






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      asked 3 hours ago









      holo

      1538




      1538






















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          You are missing the strings that both begin with three zeroes and end with two zeroes.



          And since there are five bits left that can be anything, you have $2^5=32$ of those, exactly the difference






          share|cite|improve this answer





















          • Ah. Well, I see that my problem involves not knowing that "either" also implied both in this context
            – holo
            3 hours ago










          • @holo Yeah, sometimes we use ‘either .. or’ to indicate an exclusive disjunction, but that is not always the case. For example, if I say that I would like to be either rich or happy, obviously I would not mind being rich and happy.
            – Bram28
            3 hours ago


















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          $$underbrace{2^7}_{text{begin with three zeros}}+underbrace{2^8}_{text{end with two zeros}}-underbrace{2^5}_{text{double-count}} $$






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            2 Answers
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            2 Answers
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            up vote
            3
            down vote













            You are missing the strings that both begin with three zeroes and end with two zeroes.



            And since there are five bits left that can be anything, you have $2^5=32$ of those, exactly the difference






            share|cite|improve this answer





















            • Ah. Well, I see that my problem involves not knowing that "either" also implied both in this context
              – holo
              3 hours ago










            • @holo Yeah, sometimes we use ‘either .. or’ to indicate an exclusive disjunction, but that is not always the case. For example, if I say that I would like to be either rich or happy, obviously I would not mind being rich and happy.
              – Bram28
              3 hours ago















            up vote
            3
            down vote













            You are missing the strings that both begin with three zeroes and end with two zeroes.



            And since there are five bits left that can be anything, you have $2^5=32$ of those, exactly the difference






            share|cite|improve this answer





















            • Ah. Well, I see that my problem involves not knowing that "either" also implied both in this context
              – holo
              3 hours ago










            • @holo Yeah, sometimes we use ‘either .. or’ to indicate an exclusive disjunction, but that is not always the case. For example, if I say that I would like to be either rich or happy, obviously I would not mind being rich and happy.
              – Bram28
              3 hours ago













            up vote
            3
            down vote










            up vote
            3
            down vote









            You are missing the strings that both begin with three zeroes and end with two zeroes.



            And since there are five bits left that can be anything, you have $2^5=32$ of those, exactly the difference






            share|cite|improve this answer












            You are missing the strings that both begin with three zeroes and end with two zeroes.



            And since there are five bits left that can be anything, you have $2^5=32$ of those, exactly the difference







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 3 hours ago









            Bram28

            58.3k44185




            58.3k44185












            • Ah. Well, I see that my problem involves not knowing that "either" also implied both in this context
              – holo
              3 hours ago










            • @holo Yeah, sometimes we use ‘either .. or’ to indicate an exclusive disjunction, but that is not always the case. For example, if I say that I would like to be either rich or happy, obviously I would not mind being rich and happy.
              – Bram28
              3 hours ago


















            • Ah. Well, I see that my problem involves not knowing that "either" also implied both in this context
              – holo
              3 hours ago










            • @holo Yeah, sometimes we use ‘either .. or’ to indicate an exclusive disjunction, but that is not always the case. For example, if I say that I would like to be either rich or happy, obviously I would not mind being rich and happy.
              – Bram28
              3 hours ago
















            Ah. Well, I see that my problem involves not knowing that "either" also implied both in this context
            – holo
            3 hours ago




            Ah. Well, I see that my problem involves not knowing that "either" also implied both in this context
            – holo
            3 hours ago












            @holo Yeah, sometimes we use ‘either .. or’ to indicate an exclusive disjunction, but that is not always the case. For example, if I say that I would like to be either rich or happy, obviously I would not mind being rich and happy.
            – Bram28
            3 hours ago




            @holo Yeah, sometimes we use ‘either .. or’ to indicate an exclusive disjunction, but that is not always the case. For example, if I say that I would like to be either rich or happy, obviously I would not mind being rich and happy.
            – Bram28
            3 hours ago










            up vote
            1
            down vote













            $$underbrace{2^7}_{text{begin with three zeros}}+underbrace{2^8}_{text{end with two zeros}}-underbrace{2^5}_{text{double-count}} $$






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              up vote
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              $$underbrace{2^7}_{text{begin with three zeros}}+underbrace{2^8}_{text{end with two zeros}}-underbrace{2^5}_{text{double-count}} $$






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                up vote
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                $$underbrace{2^7}_{text{begin with three zeros}}+underbrace{2^8}_{text{end with two zeros}}-underbrace{2^5}_{text{double-count}} $$






                share|cite|improve this answer












                $$underbrace{2^7}_{text{begin with three zeros}}+underbrace{2^8}_{text{end with two zeros}}-underbrace{2^5}_{text{double-count}} $$







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                answered 3 hours ago









                David Peterson

                8,55621935




                8,55621935






























                     

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