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Rearrangement of 11223344

  1. Aug 29, 2011 #1
    Find the number of rearrangements of the string 11223344 that contain no two consecutive equal digits

    So for the total rearrangements we have

    [itex]\stackrel{8!}{2!2!2!2!}[/itex]

    I was thinking I would count the total number of rearrangements that have the same digit next to each other

    So I started with the total number of rearrangements and sub tract the number of rearrangements that share the same number with the first number selected

    [itex]\stackrel{7!}{2!2!2!}[/itex](4) I multiplied by four because there are 4 numbers. From here I am confused, do you assume that you have had 2 digits in a row or do you assume you could have any rearrangement of the 6 digits?


    Im thinking the last term would be 4!, im not sure of the sign though

    So far I have

    [itex]\stackrel{8!}{2!2!2!2!}[/itex]-[itex]\stackrel{7!}{2!2!2!}[/itex](4)+............. 4!
     
    Last edited: Aug 29, 2011
  2. jcsd
  3. Aug 30, 2011 #2

    lanedance

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    yeah the problem will be accounting for like terms with double and greater repetitions, and over counting of these.

    For example if you just counted the number of a single repetition
    11223434
    would get repeated twice, when it is only one arrangement, so you need to account for this
     
  4. Aug 30, 2011 #3

    lanedance

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    so there's probably a number of ways to do it and its not always obvious which is best until you get into it, but to get started:

    counting the number of arrangements with repetitions:

    first consider a single set repeated digits
    - choose from 4 numbers
    - now choose a position in the string = 7
    - now the account for how the other 6 digits can be arranged

    But any cases that have 2,3 or 4 sets of repeated digits, will have been over-counted. I would try and calculate these explicitly and subtract...
     
    Last edited: Aug 30, 2011
  5. Aug 30, 2011 #4
    Ok I think I got it

    [itex]\stackrel{8!}{2!2!2!2!}[/itex]-[itex]\stackrel{7!}{2!2!2!}[/itex](4C1)+ [itex]\stackrel{6!}{2!2!}[/itex](4C2)-[itex]\stackrel{5!}{2!}[/itex](4C3)+4!(4C4)

    Does that seem right?
     
  6. Aug 30, 2011 #5

    lanedance

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    can you explain your method (or the meaning of) for each of the terms?
     
  7. Aug 31, 2011 #6
    The first term is the number of rearrangements of the string

    the second term is once we select the first number we have 7 left over and we can choose for 1 of the 4 numbers

    the third term is when we have 6 letters in the string ( i.e. 223344) when we choose two numbers of those next to each other we have and of the four numbers we can choose 2 different ways.

    the fourth term the number of string when there are 5 rearrangements when we choose 3 of the four numbers

    the last one is the number of ways the string keeps all of the numbers together
     
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