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Math Help - Permutation Question

  1. #1
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    Permutation Question

    Q: How many strings with seven or more characters can be formed from the letters in EVERGREEN?


    What am I suppose to do for this question? Since it asks for "seven or more", am i suppose to add up all the different possibilities?

    My answerim not sure if its right)
    [ (7!/(2!2!)) + (7!/(4!)) + (7!/(3!)) ]- For this i took out 2 E's, then 2 R's, then E and R
    +
    [ (8!/(3!2!)) + (8!/(4!)) ]- For this i took out 1 E, then 1 R.
    +
    [ 9!/(4!2!)



    Is that the correct answer? If not please correct me.

    Thanks
    Creative
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  2. #2
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    Quote Originally Posted by Creative View Post
    Q: How many strings with seven or more characters can be formed from the letters in EVERGREEN?


    What am I suppose to do for this question? Since it asks for "seven or more", am i suppose to add up all the different possibilities?

    My answerim not sure if its right)
    [ (7!/(2!2!)) + (7!/(4!)) + (7!/(3!)) ]- For this i took out 2 E's, then 2 R's, then E and R
    +
    [ (8!/(3!2!)) + (8!/(4!)) ]- For this i took out 1 E, then 1 R.
    +
    [ 9!/(4!2!)



    Is that the correct answer? If not please correct me.

    Thanks
    Creative
    Read this: http://regentsprep.org/Regents/math/...2/LpermRep.htm
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  3. #3
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    Thanks, but that did not help.
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  4. #4
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    Quote Originally Posted by Creative View Post
    Thanks, but that did not help.
    What about the formula about half way down that is relevant to your problem ....?
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  5. #5
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    This is not a hard problem but it is very time consuming.
    Just to find the number of seven strings there are many cases.
    [EEEERRN]; [EEEERRG]; [EEEERRV]; [EEEERGN]; [EEEERVN];
    [EEEERVG]; [EEEEVGN]; [EEERVGN]; [EERRVGN].
    Now count each of those cases (If I have not missed some!)

    Do the same thing for an eight string.

    The nine string is the easy case (why?)
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