Probability and counting bit string












0












$begingroup$


How many 10-bit strings are there subject to each of the following restrictions if the first two bits are the same as the last two bits?



My answer:



1) So for the first $2$ bits and the last $2$ bits we only have $2$ choices: either those $4$ bits are all 0s or all 1s. So $2$ choices.



2) Then we have $(10-2-2)$ $6$ bits left with $2$ choices so $6^2$.



Conclusion : $2 * 2^6$ or $2^7$



But, I just read that the solution is $2^8$ I don't understand why since 4 bits should be exactly the same



thanks






probability discrete-mathematics







share|cite|improve this question











$endgroup$












  • $begingroup$
    In step 1, you have two choices for each of the first two bits. Choosing the first two bits also determines the last two bits since they are the same. In step 2, you should have written $2^6$ since you have two choices for each of the six middle bits.
    $endgroup$
    – N. F. Taussig
    Nov 24 '18 at 23:36
















0












$begingroup$


How many 10-bit strings are there subject to each of the following restrictions if the first two bits are the same as the last two bits?



My answer:



1) So for the first $2$ bits and the last $2$ bits we only have $2$ choices: either those $4$ bits are all 0s or all 1s. So $2$ choices.



2) Then we have $(10-2-2)$ $6$ bits left with $2$ choices so $6^2$.



Conclusion : $2 * 2^6$ or $2^7$



But, I just read that the solution is $2^8$ I don't understand why since 4 bits should be exactly the same



thanks






probability discrete-mathematics







share|cite|improve this question











$endgroup$












  • $begingroup$
    In step 1, you have two choices for each of the first two bits. Choosing the first two bits also determines the last two bits since they are the same. In step 2, you should have written $2^6$ since you have two choices for each of the six middle bits.
    $endgroup$
    – N. F. Taussig
    Nov 24 '18 at 23:36














0












0








0





$begingroup$


How many 10-bit strings are there subject to each of the following restrictions if the first two bits are the same as the last two bits?



My answer:



1) So for the first $2$ bits and the last $2$ bits we only have $2$ choices: either those $4$ bits are all 0s or all 1s. So $2$ choices.



2) Then we have $(10-2-2)$ $6$ bits left with $2$ choices so $6^2$.



Conclusion : $2 * 2^6$ or $2^7$



But, I just read that the solution is $2^8$ I don't understand why since 4 bits should be exactly the same



thanks






probability discrete-mathematics







share|cite|improve this question











$endgroup$




How many 10-bit strings are there subject to each of the following restrictions if the first two bits are the same as the last two bits?



My answer:



1) So for the first $2$ bits and the last $2$ bits we only have $2$ choices: either those $4$ bits are all 0s or all 1s. So $2$ choices.



2) Then we have $(10-2-2)$ $6$ bits left with $2$ choices so $6^2$.



Conclusion : $2 * 2^6$ or $2^7$



But, I just read that the solution is $2^8$ I don't understand why since 4 bits should be exactly the same



thanks






probability discrete-mathematics




probability discrete-mathematics






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share|cite|improve this question













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edited Nov 24 '18 at 23:50









Key Flex

7,77441232




7,77441232










asked Nov 24 '18 at 22:40









Laura1999Laura1999

202




202












  • $begingroup$
    In step 1, you have two choices for each of the first two bits. Choosing the first two bits also determines the last two bits since they are the same. In step 2, you should have written $2^6$ since you have two choices for each of the six middle bits.
    $endgroup$
    – N. F. Taussig
    Nov 24 '18 at 23:36


















  • $begingroup$
    In step 1, you have two choices for each of the first two bits. Choosing the first two bits also determines the last two bits since they are the same. In step 2, you should have written $2^6$ since you have two choices for each of the six middle bits.
    $endgroup$
    – N. F. Taussig
    Nov 24 '18 at 23:36
















$begingroup$
In step 1, you have two choices for each of the first two bits. Choosing the first two bits also determines the last two bits since they are the same. In step 2, you should have written $2^6$ since you have two choices for each of the six middle bits.
$endgroup$
– N. F. Taussig
Nov 24 '18 at 23:36




$begingroup$
In step 1, you have two choices for each of the first two bits. Choosing the first two bits also determines the last two bits since they are the same. In step 2, you should have written $2^6$ since you have two choices for each of the six middle bits.
$endgroup$
– N. F. Taussig
Nov 24 '18 at 23:36










3 Answers
3






active

oldest

votes


















2












$begingroup$

I think you misinterpreted "first two bits are the same as the last two bits."
It is possible that the first two bits are 01, and that the last two bits are 01, for example. Thus there are $4$ possibilities for these four bits, and then multiplied by $2^6$ for the remaining six bits yields $2^8$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    If the last two bits have to be the same as the first two bits, then effectively you can only vary the first $8$ bits. So: $2^8$possible bit strings






    share|cite|improve this answer









    $endgroup$





















      1












      $begingroup$

      Each of the first $8$ bits can be either $0$ or $1$. Once the the first $8$ bits are determined, the last two
      bits must match the first two bits, so there are no remaining choices for the string. Thus, the number of strings in which the first two bits are the same as the last two bits is $2^8$






      share|cite|improve this answer









      $endgroup$













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        3 Answers
        3






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        3 Answers
        3






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        active

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        2












        $begingroup$

        I think you misinterpreted "first two bits are the same as the last two bits."
        It is possible that the first two bits are 01, and that the last two bits are 01, for example. Thus there are $4$ possibilities for these four bits, and then multiplied by $2^6$ for the remaining six bits yields $2^8$.






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          I think you misinterpreted "first two bits are the same as the last two bits."
          It is possible that the first two bits are 01, and that the last two bits are 01, for example. Thus there are $4$ possibilities for these four bits, and then multiplied by $2^6$ for the remaining six bits yields $2^8$.






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            I think you misinterpreted "first two bits are the same as the last two bits."
            It is possible that the first two bits are 01, and that the last two bits are 01, for example. Thus there are $4$ possibilities for these four bits, and then multiplied by $2^6$ for the remaining six bits yields $2^8$.






            share|cite|improve this answer









            $endgroup$



            I think you misinterpreted "first two bits are the same as the last two bits."
            It is possible that the first two bits are 01, and that the last two bits are 01, for example. Thus there are $4$ possibilities for these four bits, and then multiplied by $2^6$ for the remaining six bits yields $2^8$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 '18 at 22:50









            angryavianangryavian

            40k23280




            40k23280























                1












                $begingroup$

                If the last two bits have to be the same as the first two bits, then effectively you can only vary the first $8$ bits. So: $2^8$possible bit strings






                share|cite|improve this answer









                $endgroup$


















                  1












                  $begingroup$

                  If the last two bits have to be the same as the first two bits, then effectively you can only vary the first $8$ bits. So: $2^8$possible bit strings






                  share|cite|improve this answer









                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    If the last two bits have to be the same as the first two bits, then effectively you can only vary the first $8$ bits. So: $2^8$possible bit strings






                    share|cite|improve this answer









                    $endgroup$



                    If the last two bits have to be the same as the first two bits, then effectively you can only vary the first $8$ bits. So: $2^8$possible bit strings







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 24 '18 at 23:07









                    Bram28Bram28

                    60.5k44590




                    60.5k44590























                        1












                        $begingroup$

                        Each of the first $8$ bits can be either $0$ or $1$. Once the the first $8$ bits are determined, the last two
                        bits must match the first two bits, so there are no remaining choices for the string. Thus, the number of strings in which the first two bits are the same as the last two bits is $2^8$






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          Each of the first $8$ bits can be either $0$ or $1$. Once the the first $8$ bits are determined, the last two
                          bits must match the first two bits, so there are no remaining choices for the string. Thus, the number of strings in which the first two bits are the same as the last two bits is $2^8$






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            Each of the first $8$ bits can be either $0$ or $1$. Once the the first $8$ bits are determined, the last two
                            bits must match the first two bits, so there are no remaining choices for the string. Thus, the number of strings in which the first two bits are the same as the last two bits is $2^8$






                            share|cite|improve this answer









                            $endgroup$



                            Each of the first $8$ bits can be either $0$ or $1$. Once the the first $8$ bits are determined, the last two
                            bits must match the first two bits, so there are no remaining choices for the string. Thus, the number of strings in which the first two bits are the same as the last two bits is $2^8$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 24 '18 at 23:49









                            Key FlexKey Flex

                            7,77441232




                            7,77441232






























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