How many bit strings of length $7$ either begin with two $1's$ or end with three $1's$?












0












$begingroup$


So for the first case (beginning with 2 $1's$) there are: $2*2*2*2*2=32$ ways



Second case (end with three $1's$): $2*2*2*2=16$



And then we can just sum it $32+16=48$ different bit string of length 7 that begin with 2 $1's$ or end with 3 $1's$






combinatorics discrete-mathematics







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$endgroup$












  • $begingroup$
    Need to subtract the overlap (starts 11, ends 111), you have those twice.
    $endgroup$
    – vonbrand
    Apr 17 '14 at 0:40


















0












$begingroup$


So for the first case (beginning with 2 $1's$) there are: $2*2*2*2*2=32$ ways



Second case (end with three $1's$): $2*2*2*2=16$



And then we can just sum it $32+16=48$ different bit string of length 7 that begin with 2 $1's$ or end with 3 $1's$






combinatorics discrete-mathematics







share|cite|improve this question









$endgroup$












  • $begingroup$
    Need to subtract the overlap (starts 11, ends 111), you have those twice.
    $endgroup$
    – vonbrand
    Apr 17 '14 at 0:40
















0












0








0





$begingroup$


So for the first case (beginning with 2 $1's$) there are: $2*2*2*2*2=32$ ways



Second case (end with three $1's$): $2*2*2*2=16$



And then we can just sum it $32+16=48$ different bit string of length 7 that begin with 2 $1's$ or end with 3 $1's$






combinatorics discrete-mathematics







share|cite|improve this question









$endgroup$




So for the first case (beginning with 2 $1's$) there are: $2*2*2*2*2=32$ ways



Second case (end with three $1's$): $2*2*2*2=16$



And then we can just sum it $32+16=48$ different bit string of length 7 that begin with 2 $1's$ or end with 3 $1's$






combinatorics discrete-mathematics




combinatorics discrete-mathematics






share|cite|improve this question













share|cite|improve this question











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asked Apr 17 '14 at 0:11









athertonatherton

6561126




6561126












  • $begingroup$
    Need to subtract the overlap (starts 11, ends 111), you have those twice.
    $endgroup$
    – vonbrand
    Apr 17 '14 at 0:40




















  • $begingroup$
    Need to subtract the overlap (starts 11, ends 111), you have those twice.
    $endgroup$
    – vonbrand
    Apr 17 '14 at 0:40


















$begingroup$
Need to subtract the overlap (starts 11, ends 111), you have those twice.
$endgroup$
– vonbrand
Apr 17 '14 at 0:40






$begingroup$
Need to subtract the overlap (starts 11, ends 111), you have those twice.
$endgroup$
– vonbrand
Apr 17 '14 at 0:40












2 Answers
2






active

oldest

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1












$begingroup$

Your use of the phrase "either" is ambiguous. Can a String starting with two $1$'s not end with three $1$'s? If so, then your count is $2 * 2 * (2^{3} - 1)$ (characters 3-4 can have either $0$ or $1$, characters $5-7$ we remove the case of $111$. So there are $2^{3} - 1$ ways then to form binary strings of length $3$ that are not $111$).



For your second case, I assume that the string doesn't start with $11$. And so there are $2^{2} - 1$ ways of placing the first two characters. The last three are fixed, and so we have $2^{2} * (2^{2} - 1)$ ways of forming your string.



So by rule of sum, you add disjoint cases: $2^{5} - 2^{2} + 2^{4} - 2^{2} = 40$.



If these cases are not disjoint, then your answer is $44$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You are double-counting those strings which both begin with two $1$'s and end with three $1$'s. You need to subtract these $4$ strings from your total of $48$.
    The correct answer is thus $44$.






    share|cite|improve this answer









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      2 Answers
      2






      active

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      1












      $begingroup$

      Your use of the phrase "either" is ambiguous. Can a String starting with two $1$'s not end with three $1$'s? If so, then your count is $2 * 2 * (2^{3} - 1)$ (characters 3-4 can have either $0$ or $1$, characters $5-7$ we remove the case of $111$. So there are $2^{3} - 1$ ways then to form binary strings of length $3$ that are not $111$).



      For your second case, I assume that the string doesn't start with $11$. And so there are $2^{2} - 1$ ways of placing the first two characters. The last three are fixed, and so we have $2^{2} * (2^{2} - 1)$ ways of forming your string.



      So by rule of sum, you add disjoint cases: $2^{5} - 2^{2} + 2^{4} - 2^{2} = 40$.



      If these cases are not disjoint, then your answer is $44$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        Your use of the phrase "either" is ambiguous. Can a String starting with two $1$'s not end with three $1$'s? If so, then your count is $2 * 2 * (2^{3} - 1)$ (characters 3-4 can have either $0$ or $1$, characters $5-7$ we remove the case of $111$. So there are $2^{3} - 1$ ways then to form binary strings of length $3$ that are not $111$).



        For your second case, I assume that the string doesn't start with $11$. And so there are $2^{2} - 1$ ways of placing the first two characters. The last three are fixed, and so we have $2^{2} * (2^{2} - 1)$ ways of forming your string.



        So by rule of sum, you add disjoint cases: $2^{5} - 2^{2} + 2^{4} - 2^{2} = 40$.



        If these cases are not disjoint, then your answer is $44$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          Your use of the phrase "either" is ambiguous. Can a String starting with two $1$'s not end with three $1$'s? If so, then your count is $2 * 2 * (2^{3} - 1)$ (characters 3-4 can have either $0$ or $1$, characters $5-7$ we remove the case of $111$. So there are $2^{3} - 1$ ways then to form binary strings of length $3$ that are not $111$).



          For your second case, I assume that the string doesn't start with $11$. And so there are $2^{2} - 1$ ways of placing the first two characters. The last three are fixed, and so we have $2^{2} * (2^{2} - 1)$ ways of forming your string.



          So by rule of sum, you add disjoint cases: $2^{5} - 2^{2} + 2^{4} - 2^{2} = 40$.



          If these cases are not disjoint, then your answer is $44$.






          share|cite|improve this answer









          $endgroup$



          Your use of the phrase "either" is ambiguous. Can a String starting with two $1$'s not end with three $1$'s? If so, then your count is $2 * 2 * (2^{3} - 1)$ (characters 3-4 can have either $0$ or $1$, characters $5-7$ we remove the case of $111$. So there are $2^{3} - 1$ ways then to form binary strings of length $3$ that are not $111$).



          For your second case, I assume that the string doesn't start with $11$. And so there are $2^{2} - 1$ ways of placing the first two characters. The last three are fixed, and so we have $2^{2} * (2^{2} - 1)$ ways of forming your string.



          So by rule of sum, you add disjoint cases: $2^{5} - 2^{2} + 2^{4} - 2^{2} = 40$.



          If these cases are not disjoint, then your answer is $44$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Apr 17 '14 at 0:18









          ml0105ml0105

          11.5k21538




          11.5k21538























              1












              $begingroup$

              You are double-counting those strings which both begin with two $1$'s and end with three $1$'s. You need to subtract these $4$ strings from your total of $48$.
              The correct answer is thus $44$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                You are double-counting those strings which both begin with two $1$'s and end with three $1$'s. You need to subtract these $4$ strings from your total of $48$.
                The correct answer is thus $44$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  You are double-counting those strings which both begin with two $1$'s and end with three $1$'s. You need to subtract these $4$ strings from your total of $48$.
                  The correct answer is thus $44$.






                  share|cite|improve this answer









                  $endgroup$



                  You are double-counting those strings which both begin with two $1$'s and end with three $1$'s. You need to subtract these $4$ strings from your total of $48$.
                  The correct answer is thus $44$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Apr 17 '14 at 0:14









                  jwsiegeljwsiegel

                  1,333510




                  1,333510






























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