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Math Help - Combinations/ Permutation

  1. #1
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    Exclamation Combinations/ Permutation

    i have discovered a question that i am stuck on again. Help would be appreciated!

    In how many ways can you arrange the word BOOKKEEPER without having identical letters (such as OO) beside each other?
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  2. #2
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    I am going to be brave here and give it a go. Perhaps if I'm overcounting or undercounting, someone will be along.

    Start by arranging the R,P, and B with spaces in between and on the ends.

    _B_P_R_

    This gives us 4 positions to choose 3 and 3! ways to arrange the B,P,R.

    After some head-scratching and trying to account for the different cases.

    We get 3!C(4,3)C(7,2)C(9,2)=18144
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  3. #3
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    The trouble with that answer is that it would never count the following:
    EOEBOKEPRK.
    That is, it may not count cases in which two repeats are separated by another repeater as in the example above.
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  4. #4
    Eater of Worlds
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    Oh well, this isn't an easy problem.

    We start with 10 letters.

    We arrange the P,R,B in a line with the gaps. This leaves 4 places we can choose 3 to place the E. This gives the 24 ways as per the last post.

    Then, we have 6 letters and we arrange them with gaps in between on the ends. This gives 7 positions we can choose 2 to place the K. C(7,2)=21

    Now we have 8 letters and we arrange with gaps in between and on the ends. This gives 9 places and we can choose 2 to place the O. C(9,2)=36 ways.

    We have 24*21*36=18144.

    That seems OK, but if I am missing something then..............
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  5. #5
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    I understand how you got the above answer. I also agree that it is a lower bound for the correct answer.

    But once we have placed the E’s there is never a way to get a string containing “EOE…” or a string containing “EKE…” because there is never a gap between two E’s (there are not together to begin with). If we place the O’s first then we will never get a string containing “OEO…” or a string containing “OKO…” for the same reason.

    I am willing to wager that who ever wrote this question did not think it though. If there were just two E’s, then something like to old couples problem could be used.
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  6. #6
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    It always surprises me how the most experienced counter can be blinded by a false start.
    Suppose that O is the set of rearrangements in which the two Os are together; K is the set of rearrangements in which the two Ks are together; and E is the set of rearrangements in which the three Es are together.
    We want to count the number \left( {O^c  \cap K^c  \cap E^c } \right) = \left( {O \cup K \cup E} \right)^c .

    \left| O \right| = \left| K \right| = \frac{{9!}}{{\left( {2!} \right)\left( {3!} \right)}}\,\& \,\left| E \right| = \frac{{8!}}{{\left( {2!} \right)^2 }}.

    \left| {O \cap K} \right| = \frac{{8!}}{{3!}}\,\& \,\left| {O \cap E} \right| = \left| {K \cap E} \right| = \frac{{7!}}{{2!}}

    \left| {O \cap K \cap E} \right| = 6!

    Can you put that all together?
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