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Permutations with possibly repeating letters

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  1. Jan 20, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the number of distinct three letter words obtainable from the letters A,B,C,D,E in which E may occur 0, 1 or 2 times but the rest may occur only once.
    2. Relevant equations
    Number of combinations when picking r objects from n possibly objects in which one object is repeated m times
    N = n! / (m! (n-r)!
    3. The attempt at a solution
    I have never had a good instinct for permutations, I can plug in a formula to answer basic questions but never had a good enough understanding for it to solve more advanced problems (This is why I have dug out my old maths textbook to try and correct this )

    One thing that is especially confusing for me is this.
    With one E there are 5P3 = 5!/2! = 60 combinations (but this already includes ABC, ABD, ... etc i.e. all of the possible combinations with zero E's ?).
    If I found the number of permutations for 3 letter words with 2E's [6!/(2!3!)] shouldn't this also include ABE etc i.e. all the permutations with 1E? Yet the answer is 60 so clearly it doesn't. I think if I could resolve my misunderstanding if this part the fog would lift for me.

    Please let me know if anything is unclear and I will make some edits
     
  2. jcsd
  3. Jan 20, 2017 #2

    PeroK

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    To build your understanding, I suggest you try this problem without using the combination or permutation formulas. Try to count them more directly and find your own shortcuts.

    Hint: Calculate separately the cases of 0, 1 or 2 E's.
     
  4. Jan 20, 2017 #3

    haruspex

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    Yes. 0 or 1 Es can be handled together because that makes E behave the same as the other letters.
    Exactly 2 Es? I make it rather less.
    I think you are asking whether the way you arrived at 6!/(2!3!) includes ABE etc. I cannot tell since you do not explain it.
     
  5. Jan 21, 2017 #4
    Thank you, yes you have understood me correctly! That is my question. I thought that 6!/(2!3!) was basically equivalent to finding distinct permutations of ABCDEE (which would include ABE etc but clearly not). What does 6!/(2!3!) tell you the number of?
     
  6. Jan 21, 2017 #5

    haruspex

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    You correctly found 60 with 0 or 1 Es. 6!/3! (=120) would count words with up to 2 Es if we regarded the 2 Es as distinct. You divided by 2 because the 2 Es are interchangeable. That gives you the number with exactly 2 Es plus half the number with 0 or 1 Es (totalling 60 again). So... how many have exactly 2 Es?
     
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