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2nd combinatorial problem

  1. Jul 18, 2007 #1
    1. The problem statement, all variables and given/known data
    Decide in how many ways the letters of the word ABRACADABRA can be arranged in a row if C, R and D are not to be together.


    2. Relevant equations
    The number of ways of arranging n objects which include 'a' identical objects of one type, 'b' identical objects of another type,.... is
    n!/(a!b!....)


    n objects divided into m groups with each group having G1, G2, ..., Gm objects respectively has m! * G1! * G2! * ... *Gm!



    3. The attempt at a solution
    A: 5
    B: 2
    R: 2
    C: 1
    D: 1

    11 letters in total.

    There are a few identical letters so the total number of ways of permuting the objects accounting for the identical letters is 11!/(5!2!2!)

    The total number of ways of arranging the letters such that C, R and D are together is:
    8!4!/(5!2!2!) as there are 8 groups, with one group containing 4 letters. However given the identical letters, we divide by the same number as above.

    So (11!-8!4!)/(5!2!2!)=81144

    The answers suggested 78624

    I can't see what is wrong with my reasoning.
     
  2. jcsd
  3. Jul 18, 2007 #2
    total ways of arranging = 11! / (5!.2!.2!)

    exceptions=> ways in which CRD are together = (9!.3!) / (5!.2!.2!)

    why (9!.3!) / (5!.2!.2!) ? Consider CRD to be one grp and the rest 9 to be another. Then 9! ways or arranging those letters, 3! ways of arranging CRD, 5 A's are common, 2 B's. Another 2! in Dr. because there are two R's, and it doesnt matter which R is in the word CRD
     
  4. Jul 18, 2007 #3
    Nice one f(x). I see now.
     
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