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Math Help - form a binary string

  1. #1
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    form a binary string

    Count the ways that a 100 length binary string can be formed if the times that the digit zero appears is an even number and the times that the digit one (1) appears is an even number as well. (Note: in the even numbers zero is inluded.)

    Dont get it.
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  2. #2
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    Quote Originally Posted by milagros View Post
    Count the ways that a 100 length binary string can be formed if the times that the digit zero appears is an even number and the times that the digit one (1) appears is an even number as well. (Note: in the even numbers zero is inluded.)

    Dont get it.
    How many way can a binary string of length 100 be formed? (The binary string consists ONLY of 1's & 0's.)

    Then take half.
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  3. #3
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    Quote Originally Posted by aidan View Post
    How many way can a binary string of length 100 be formed? (The binary string consists ONLY of 1's & 0's.)

    Then take half.
    Must have even zeros and even ones. Example for four length string:

    1001
    1100
    1010
    0011
    0101
    ......

    etc
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  4. #4
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    Quote Originally Posted by milagros View Post
    ...Example for four length string:
    1001
    1100
    1010
    0011
    0101
    ...etc
    Sorry my post wasn't meaningful to you.
    A better example:
    How many ways can you have a string of 1's & 0's of length four?
    Code:
           bit
           cnt
    0000   0      1
    0001   1      2
    0010   1      3
    0011   0      4
    0100   1      5
    0101   0      6
    0110   0      7
    0111   1      8
    1000   1      9
    1001   0      10
    1010   0      11
    1011   1      12
    1100   0      13
    1101   1      14
    1110   1      15
    1111   0      16
    You should say 'sixteen' ( since they are also sequenced ).
    How many had an even number of 1's in the string.
    If you said 'half' or 'eight' you are onto something.

    Here is another example:
    How many ways can you have a string of 1's & 0's of length six?
    Code:
             bit
             cnt 
    000000   0      1
    000001   1      2
    000010   1      3
    000011   0      4
    000100   1      5
    000101   0      6
    000110   0      7
    000111   1      8
    001000   1      9
    001001   0      10
    001010   0      11
    001011   1      12
    001100   0      13
    001101   1      14
    001110   1      15
    001111   0      16
    010000   1      17
    010001   0      18
    010010   0      19
    010011   1      20
    010100   0      21
    010101   1      22
    010110   1      23
    010111   0      24
    011000   0      25
    011001   1      26
    011010   1      27
    011011   0      28
    011100   1      29
    011101   0      30
    011110   0      31
    011111   1      32
    100000   1      33
    100001   0      34
    100010   0      35
    100011   1      36
    100100   0      37
    100101   1      38
    100110   1      39
    100111   0      40
    101000   0      41
    101001   1      42
    101010   1      43
    101011   0      44
    101100   1      45
    101101   0      46
    101110   0      47
    101111   1      48
    110000   0      49
    110001   1      50
    110010   1      51
    110011   0      52
    110100   1      53
    110101   0      54
    110110   0      55
    110111   1      56
    111000   1      57
    111001   0      58
    111010   0      59
    111011   1      60
    111100   0      61
    111101   1      62
    111110   1      63
    111111   0      64
    Obviously, 64.
    Maybe not so obvious is this:  2^6 = 64 .
    How many had an even count of 1's (or zeroes) in the string?
    If you count them you will find 32.
    Half of 64 (or 2^6) is 32 (or 2^5).


    The FINAL EXAM!
    Count the ways that a 100 length binary string can be formed if the times that the digit zero appears is an even number and the times that the digit one (1) appears is an even number as well. (Note: in the even numbers zero is inluded.)
    How many ways can you have a string of 1's & 0's of length 100?
    Recall:
    a string length of 4 has 2^4 ways
    a string length of 6 has 2^6 ways.
    That should be easy.

    The difficult part:
    Now take half of that.
    Recall: half of 2^6 is 2^5

    .
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  5. #5
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    Quote Originally Posted by aidan View Post
    Sorry my post wasn't meaningful to you.
    A better example:
    How many ways can you have a string of 1's & 0's of length four?
    Code:
           bit
           cnt
    0000   0      1
    0001   1      2
    0010   1      3
    0011   0      4
    0100   1      5
    0101   0      6
    0110   0      7
    0111   1      8
    1000   1      9
    1001   0      10
    1010   0      11
    1011   1      12
    1100   0      13
    1101   1      14
    1110   1      15
    1111   0      16
    You should say 'sixteen' ( since they are also sequenced ).
    How many had an even number of 1's in the string.
    If you said 'half' or 'eight' you are onto something.

    Here is another example:
    How many ways can you have a string of 1's & 0's of length six?
    Code:
             bit
             cnt 
    000000   0      1
    000001   1      2
    000010   1      3
    000011   0      4
    000100   1      5
    000101   0      6
    000110   0      7
    000111   1      8
    001000   1      9
    001001   0      10
    001010   0      11
    001011   1      12
    001100   0      13
    001101   1      14
    001110   1      15
    001111   0      16
    010000   1      17
    010001   0      18
    010010   0      19
    010011   1      20
    010100   0      21
    010101   1      22
    010110   1      23
    010111   0      24
    011000   0      25
    011001   1      26
    011010   1      27
    011011   0      28
    011100   1      29
    011101   0      30
    011110   0      31
    011111   1      32
    100000   1      33
    100001   0      34
    100010   0      35
    100011   1      36
    100100   0      37
    100101   1      38
    100110   1      39
    100111   0      40
    101000   0      41
    101001   1      42
    101010   1      43
    101011   0      44
    101100   1      45
    101101   0      46
    101110   0      47
    101111   1      48
    110000   0      49
    110001   1      50
    110010   1      51
    110011   0      52
    110100   1      53
    110101   0      54
    110110   0      55
    110111   1      56
    111000   1      57
    111001   0      58
    111010   0      59
    111011   1      60
    111100   0      61
    111101   1      62
    111110   1      63
    111111   0      64
    Obviously, 64.
    Maybe not so obvious is this:  2^6 = 64 .
    How many had an even count of 1's (or zeroes) in the string?
    If you count them you will find 32.
    Half of 64 (or 2^6) is 32 (or 2^5).


    The FINAL EXAM!


    How many ways can you have a string of 1's & 0's of length 100?
    Recall:
    a string length of 4 has 2^4 ways
    a string length of 6 has 2^6 ways.
    That should be easy.

    The difficult part:
    Now take half of that.
    Recall: half of 2^6 is 2^5

    .
    Thanks for being so analytical . Yesterday I thought exactly ( I thing ) the same. If I am correct you mean that my answer is 2^100 / 2 = 2^99 right?

    I made it also with examples yesterday and I thing that I am right ?? And something important: I must express the result with a GENERATOR FUNCTION. Do

    you know how to do this?

    Thanks again.
    Last edited by milagros; October 22nd 2009 at 10:47 PM.
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  6. #6
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    Think I got it thanks.
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