A question involving permutations and probability.

[missiledragon]

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

 

[Bacle2]

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.

 

[HallsofIvy]

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