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A question involving permutations and probability.

  1. Apr 1, 2013 #1
    So, (this is not homework), I have been stuck on a question regarding probability for a while now and am clueless of how to continue. The question is:

    The letters A, E, I, P, Q, and R are arranged in a circle. Find the probability that at least 2 vowels are next to one another.

    My attempt: Alright, since one letter is fixed, that leaves us with 5 letters to arrange. I'm going to fix the consonant in this case, since it is particularly easier for me :P. The remaining spots (5 left) can be arranged 3!*2! (remaining vowels and consonants respectively).

    I'm lost here though, so any help on what to do/correct me would help. Thanks! :D
  2. jcsd
  3. Apr 1, 2013 #2


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    I would consider something like this:

    You have 3 vowels {A,E,I} and 3 consonants {P,Q,R}

    You want arrangements of the types:

    1) V1, C1 , V2 ,C2, V3, C3 , where Vi is a vowel, Ci a consonant, or:

    2) C1,V1 ,C2 , V2, C3, V3

    To guarantee that there are no consecutive vowels. Every other arrangement will contain consecutive

    vowels. Then you have to put the arrangement in a circle and consider the ones that are equal as

    permutations, e.g:

    A P E Q I R is the same as P E Q I R A , when put into a circle, and so is E Q I R A P , etc.
  4. Apr 1, 2013 #3


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    Wow! missledragon waited 13 whole minutes before bumping! That may be a new record.
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