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Combinatorics - Specific Outcomes

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    An experiment has 3 outcomes. If the experiment is performed 4 times, how many sets of outcomes exist in which exactly two of the experiments had the same outcome.


    3. The attempt at a solution

    From what I understand there is a 4-element set with each element having 3 possible values. So there are 34 = 81 such sets. The question is asking how many sets contain elements where exactly two of the elements are equal. I'm having trouble understanding how to count these. I imagine starting with 4C2 to get the number of ways to get two of the elements to fit a specific criteria, but I don't know how to continue.

    Is there a general formula or reasoning process I am missing here?
     
  2. jcsd
  3. Mar 9, 2013 #2

    Bacle2

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    You're right so far; only thing left is to specify which of the 3 outcomes was repeated when the experiment was done. So you first select the experiment orders in which the outcome was repeated, e.g., (1,2), (1,3) , etc. and then you decide which of the possible outcomes was repeated.
     
  4. Mar 9, 2013 #3
    Awesome!

    So from what I understand:

    4C2 gives the number of ways the experiment could repeat itself (the number of ways certain elements in the set could meet specific criteria). Then I multiply 4C2 by the possible outcomes for the other two elements. Since one of the three outcomes has occurred the max number of times we are looking for, then each of the other two outcomes of the experiment only has two possible outcomes. So the answer is 4C2(2)(2)=24.

    So if I wanted the number of sets in which the experiment repeated 3 times I could use this calculation: 4C3(2).

    Thanks for the help!
     
  5. Mar 9, 2013 #4

    vela

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    If it's the first two experiments that have the same result, you have these possibilities:

    AABC
    AACB
    BBAC
    BBCA
    CCAB
    CCBA

    Can you see how you'd be able to calculate there are 6 possibilities?

    EDIT: I see you posted while I was writing. You actually missed Bacle2's point, but you did pick up on what Bacle2 missed, sort of. You're undercounting and overcounting right now.
     
  6. Mar 9, 2013 #5
    Dang. Can you help me understand what I missed?

    I've been having a lot trouble thinking through the logic of these types of problems.

    4C2=6 gives number of ways the experiment could repeat outcomes. 4C2(2)(2)=24 gives the number of sets in which two outcomes repeat, but this is undercounting?

    So, was I forgetting about how each outcome could be repeated? So, would it be 4C2(2)(2)(3)=72?
     
  7. Mar 9, 2013 #6

    vela

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    You can't use 2x2 to count the number of outcomes for the non-matching experiments, otherwise you include the possibility of something like AABB. That's not an event where exactly two experiments have the same outcome. So in that sense, you're overcounting.

    You're undercounting in the sense that you haven't taken into account what Bacle2 said.

    There are 6 possibilities for each of the 4C2 outcomes (so the answer is 36). The way it breaks down is into (4C2)x2x3. Try to understand where each of those last two factors comes from.
     
    Last edited: Mar 9, 2013
  8. Mar 9, 2013 #7
    Ooooh, it's starting to make sense. Thanks for your patience.

    I'll try again.

    4C2 gives the number ways that two elements could meet some criteria. There are 3 values that these two elements could have, and 2 values for the next and 1 for the last. Is that why it is (4C2)x2x3?
     
  9. Mar 9, 2013 #8

    Bacle2

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    Think of the ways in which you can have a pair of events be equal to each other; they can all be equal to....

    This was the point I was trying to make initially, but I was not clear-enough.
     
  10. Mar 9, 2013 #9
    Sorry for misunderstanding.

    So in the example above, there are 4C2 pairs of events which can be equal in 3 ways, then accounting for the last two elements having 2 and 1 outcomes we get (4C2)(3)(1)(2).

    Is that it?

    This is where I get confused with most problems in combinatorics. It seems like there are several correct interpretations, and I often mix those together to end up with a wrong interpretation... any advice on how to get this straight in my head?
     
  11. Mar 9, 2013 #10

    vela

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    Practice. :smile:
     
  12. Mar 9, 2013 #11
    Then practice it is :)

    Thanks for the help today.
     
  13. Mar 9, 2013 #12

    Bacle2

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    Oh, no,please, don't blame yourself; I really meant it, I am not always as clear as I should be. Adding to Vela's comment, try doing a few problems in great level of detail. Try, e.g., calculating the expected return for some lottery, or the probability of n people winning when x people play,or the probability of getting different hands in poker, etc. and post it here.
     
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