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Permutation and combination problems.

  1. Feb 25, 2012 #1
    1. The problem statement, all variables and given/known data

    I forget the exact expression of the questions. But the related details are exhaustive here.
    it's about permutation and combination. By the way, I am not student, i am looking for explanation and understanding, not answers..

    1. There are 6 people, namely A, B, C, D, E, F in a row, how many ways of arrangement are there such that person A,B,C must be separated

    2. Relevant equations

    No equation given


    3. The attempt at a solution

    I especially paid attention to the word "must be separate", then I first count the number of ways that A,B and C "must be grouped" together. which result in 3! (arrangement within A,B,C) and then multiply 4! (taken ABC as one object).

    Then for 6 people randomly arranged in order there are 6!. It's easy.

    so the solution i think should be 6! - 3! x 4!

    however, that was wrong. the solution is 3! x 4!

    I feel so hard understanding why.

    I don't know what they mean by "must be separate" here, how about, for example, A E F B C D? when B and C are combined while A is separate from them. Did I forgot to minus this possibility?

    The answer is actually 144 but I get no idea about it.


    I am not really clever, please explain in a simple way. Thanks people. Thanks people
     
  2. jcsd
  3. Feb 25, 2012 #2

    tiny-tim

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    hi kenny! :wink:
    yes, that's right :smile:

    "separate" means no two are together

    here's another way to do it:

    place A B and C first, with two gaps in between (how many ways are there of doing that?)

    then place D E and F (how many ways are there of doing that? you know that at least one of them has to be in the first gap)​

    have another go! (both ways) :smile:
     
  4. Feb 25, 2012 #3
    Another explanation:
    1. Arrange D, E, F as shown -D-E-F- , there are x ways to do that.
    2. Now there are 4 positions (each denoted by a dash) to place A, B, C at.
    First choose 3 positions from 4, there are y ways to do that.
    Then make an arrangement among 3 peoples, this could be done in z ways.
    3. Multiply things out and you'll get the answer.
     
  5. Feb 25, 2012 #4

    tiny-tim

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    i wonder! :rolleyes:

    can you calculate how many different ways are there of solving this question? :biggrin:
     
  6. Feb 25, 2012 #5
    i've worked out a formula for this problem, but is the formula correct?


    3!x3! + 3!x3! + 3!x3! + 3!x3! = 144
     
  7. Feb 25, 2012 #6

    tiny-tim

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    well, the result is correct :smile:

    but what's your justification for it? :confused:
     
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