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Probability: arrangements with identical elements

  1. Sep 15, 2007 #1
    1) If we have 5 letters: A, A, B, C, D, in how many ways (arrangements) can we form a 3-letter "word"?

    How can I calculate this? The 2 A's seem to make things very complicated...and I have no clue how to do it...

    It's arrangements, so I think permutation will be used.

    Also, there are two identical "A"s, so some arrangements will be double counted or so...

    Denote 1st A=A1 and 2nd A=A2
    A1BC and A2BC are counted as 1 "word" because actually A1=A2

    Thanks for helping!
     
  2. jcsd
  3. Sep 15, 2007 #2

    HallsofIvy

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    Yes, that is a little complicated! If you do as you say, and label the As as A1 and A2, there 5 distinct letters and so 5!/3!= 20 distinct words. However, you are correct that that is over counting because, in fact, some words are "the same". Try this: how many words are there that contain no As? That's easy- with no As, you just have BCD and there are 3! of them (the different permutations of BCD). How many different three letter words are there if you include only one A? That's again easy because now you are just looking at {A, B, C, D} with only one A. Finally, how many three letter words are there if you include both As? AAB, ABA, BAA, AAC, ACA, CAA, AAD. ADA, DAA: that's all the possibliities and there are exactly 9 of them: the number of three letter distinct words that can be formed from A, A, B, C, D is the sum of those three numbers. (It is surprizingly close to 20!)
     
  4. Sep 16, 2007 #3
    "If you do as you say, and label the As as A1 and A2, there 5 distinct letters and so 5!/3!= 20 distinct words" <---shouldn't it be 5 x 4 x 3 = 60 words in this case?


    So is the answer to #1 going to be 3! + (4 x 3 x 2) + the number of possible arrangments with exactly 2 A's ?
    Also, is there any way to find the number of possible arrangements with exactly 2 A's without having to LIST them all out?


    Thanks!
     
  5. Sep 16, 2007 #4

    learningphysics

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    careful 4*3*2 isn't the number of arrangements with exactly 1 A... you're including possibilities where A isn't picked.
     
  6. Sep 16, 2007 #5
    Oh no...then how can I find the number of arrangements with exactly 1 A? I am so confused...
     
  7. Sep 16, 2007 #6

    learningphysics

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    You have 3 positions for the 1A... so that's 3 possibilities... for each of those possibilities, you have 2 slots to put in B,C or D...

    so you 3*P(3,2)

    or in other words 3*(3*2)
     
  8. Sep 16, 2007 #7
    Sorry, I'll have to reconsider my solution.
     
    Last edited: Sep 16, 2007
  9. Sep 17, 2007 #8
    Thanks, I understand this part now!

    And my final question is:
    Is there any way to calculate the number of possible arrangements correctly with exactly 2 A's without having to LIST them all out? (because it's very easy to miss something by listing, in my opinion)
     
  10. Sep 17, 2007 #9

    learningphysics

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    Yes, it's very similar to analyzing the 1 A case. For 2A's... you have 3 possible positions for the 3rd letter. And for each of those positions, you can make 3 choices for the 3rd letter... B, C or D. So you have 3*C(3,1) = 3*3 = 9
     
  11. Sep 17, 2007 #10
    Nice, now I understand it perfectly. Thank you!
     
  12. Sep 18, 2007 #11
    Combinations with identical elements:
    2) You have 2 blue balls, 3 yellow balls, and 3 red balls in a box. If you randomly take 3 balls from the box, what is the probability that you have 1 ball of each color? (assume order does not matter)


    This is even crazier, how can I even get a starting point for this problem? I know the definition of combination, but I still have absolutely no clue on solving this problem...

    Any help is appreciated!
     
  13. Sep 18, 2007 #12

    learningphysics

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    Give it a shot yourself. Make a guess... we'll help you along.
     
  14. Sep 18, 2007 #13
    My guess:
    2C1 x 3C1 x 3C1
    --------------------
    (8 x 7 x 6) / (2! 3! 3!)

    But I don't think it's right...
     
  15. Sep 18, 2007 #14

    learningphysics

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    I think you're actually quite close... your numerator is correct. it is the number of ways to choose a blue ball, yellow ball and a red ball.

    So all you need in the denominator is the total number of ways of choosing 3 balls out of 8.
     
    Last edited: Sep 18, 2007
  16. Sep 18, 2007 #15
    2) So would it be
    2C1 x 3C1 x 3C1
    ---------------- ?
    8C3

    Would this in the denominator account for the identical balls?



    Also, do I have to multiply the numerator by 6? (since there are 5 mutually exclusive events that can give one ball of each color when order doesn't matter.

    The 6 mutually exclusive events are:
    RBY (first ball red, second ball blue, third ball yellow)
    RYB
    BYR
    BRY
    YBR
    YRB


    Thanks!
     
  17. Sep 18, 2007 #16

    learningphysics

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    Yes, this would work.

    We aren't using permutations in the numerator or denominator. We're using combinations. If we were using permutations, then in the numerator we'd have 2C1 x 3C1 x 3C1 x 3! and in the denominator we'd have P(8,3). This would also work... it gives the same answer.

    What wouldn't work is having permutations in the numerator and cominbations in the denominator... or combinations in the numerator and permutations in the denominator.

    Using permutations in both the numerator and denominator... or combinations in both the numerator and denominator... will work.
     
  18. Sep 20, 2007 #17
    Thanks! I am sort of getting it now.


    But what if I change the question to:
    You have 2 blue balls, 3 yellow balls, and 3 red balls in a box. If you randomly take 3 balls from the box, what is the probability that you have the first ball red, second ball yellow, and third ball blue? (order DOES matter)
     
  19. Sep 20, 2007 #18

    learningphysics

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    Then it would be (2*3*3)/P(8,3)
     
  20. Sep 21, 2007 #19
    Why?

    Why not
    2P1 x 3P1 x 3P1
    ---------------- ?
    8P3
     
  21. Sep 21, 2007 #20

    learningphysics

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    Same thing. 2P1 = 2. 3P1 = 3.
     
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