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Combination question

  1. May 7, 2014 #1

    joshmccraney

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    1. The problem statement, all variables and given/known data
    how many ways can ##a,b,c,d,e,f,g,h,i,j## (first ten letters of the alphabet) be interchanged if ##a,b,c## must be adjacent and if ##d## cannot be touching the ##a## nor can it touch the ##b##?

    3. The attempt at a solution
    to start, i figure we can take the total number of ways that ##a,b,c## without the ##d## restriction and then subtract the ways the ##d## mingles with the ##a,b##. thus, we have (i think)
    $$3!8! - 3 \times 3!7!$$

    thanks for the help!
     
  2. jcsd
  3. May 7, 2014 #2

    joshmccraney

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    actually, i think i need to make some cases, thus would it be:
    $$3!8! - (3\times 2!2! + 4 \times 2!)7!$$
    where the first case is when we have the ##a,b## adjacent and the second case we have the ##c## always in the middle.

    what do you think?
     
  4. May 9, 2014 #3

    haruspex

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    I think you're still subtracting too much. The 7! factor is fine, of course.
    Remember you should only be subtracting cases where a, b and c are an adjacent trio. That doesn't leave many illegal places for the d.
     
  5. May 10, 2014 #4

    joshmccraney

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    hmmmm how about $$3!8! - (2 \times 2!2! + 4 \times 2!)7!$$ if the ##a,b## are adjacent we have ##2!## spots for them and then ##2!## spots for them and the ##c##. this then leaves us with ##2## spots for the ##d## to still touch the ##a,b##.

    we also have the situation where the ##a,b## are not adjacent, and the ##c## is in the middle. thus, there are ##2!## ways to arrange the ##a,b## around the ##c##. however, there are now ##4## spots we can place the ##d##.

    without the ##7!## this leaves only 16 arrangements. that sounds good or too much?
     
  6. May 10, 2014 #5

    haruspex

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    After placing the c, I only see one spot for d to go to touch one of a, b.
    I only see two for the d.
    Placing d must not disrupt the existing assumed adjacencies.
     
  7. May 10, 2014 #6

    joshmccraney

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    Abdc
    Badc
    Dabc
    Dbac
    Cabd
    Cbad
    Cdab
    Cdba

    I think there's 8, right?
     
  8. May 10, 2014 #7

    haruspex

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    Your 3!8! counts only cases where a, b, c form an adjacent triple. Abdc is not one of those, so it doesn't need to be subtracted.
     
  9. May 10, 2014 #8

    joshmccraney

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    great call! i don't know how i missed this! just for completeness, is it $$3!8! - (2 \times 2! + 2 \times 2!)7!$$
    thanks for your help!
     
  10. May 11, 2014 #9

    haruspex

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    Yep.
     
  11. May 11, 2014 #10
    Yes. That answer is correct.
     
  12. May 11, 2014 #11

    adjacent

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    hmm. A soldier of me just reported that my name has been mentioned in this thread ##7\times## :approve:
    Anyway,glad that you solved a question involving my name. :wink:
     
  13. May 11, 2014 #12

    joshmccraney

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    It was anything but easy! Thank goodness for pf!
     
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