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Math Help - Permutation involving restriction

  1. #1
    Senior Member Stroodle's Avatar
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    Permutation involving restriction

    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.

    My text says the answer is 78,624 ways, but I don't know how they arrived at this.

    Thanks for your help!
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  2. #2
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    Instead of considering AB and CDR as seperate letters, consider them as 1 and 0 (not actual numbers). And then consider the possible ways you can arrange the four 0's about 11 spots (drawing a diagram might help).
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  3. #3
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    Quote Originally Posted by Stroodle View Post
    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.

    My text says the answer is 78,624 ways, but I don't know how they arrived at this.
    There are several problems with the wording of this question.
    What do you think "if C, R and D are not to be together" means?
    Does it mean CRD in that order? Or is it any order like DCR or RDC?
    Problem: There are two R's.

    I spent some time trying to find a reading that gives the given answer.
    But I cannot. I may be wrong or I don't understand what "if C, R and D are not to be together" means.
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  4. #4
    Senior Member Stroodle's Avatar
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    Yep. I can't work it out either. I copied the question word for word from my text book, but as you can see my text isn't very good.

    I just got a copy of the solutions today and it says:

    \frac{11!}{5!2!2!}-\frac{9!3!}{5!2!2!}=78624

    But I still can't work out where they got the \frac{9!3!}{5!2!2!} from.
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  5. #5
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    I assumed that "if C, R and D are not to be together" meant that those three letters must not occur consecutively (in any order). With that interpretation, I made the answer 74,424, as follows.

    Total number of arrangements of letters A(5), B(2), R(2), C, D is \frac{11!}{5!2!2!} = 83,160.
    For each of 3! orderings of C,D,R, number of arrangements of the collection A(5), B(2), R, {CDR} is \frac{9!}{5!2!}, for a total of 6\times1512 = 9072.

    So we must subtract 9072 from 83,160. But there has been some double counting: the sequence RCDR has been counted as (RCD)R and also as R(CDR), and similarly for RDCR. The number of arrangements of the set A(5), B(2), {RCDR} is \frac{8!}{5!2!}, so we must add 2\times168=336 to get the final result 83,160 9072 + 336 = 74,424.
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  6. #6
    Senior Member Stroodle's Avatar
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    Awesome. That makes sense. Thanks heaps for that. And thanks for spending the time to help on on such an ambiguous question Plato. Really appreciate it.

    Strange that both the back of the textbook and the separate worked solutions have the same incorrect answer...
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