I would calculate, using the fact that there are
ways to have A, E, I, O, U in that order.
For each of these, there are 6! ways to arrange the remaining letters.
Hence, the probability is
I just wrote a practice midterm for a probability course. There was no answer key and the person who "marked" the exam and I were unsure about one of the questions.
"What is the probability that in a random permutation of the nine letters A, E, I, O, U, C, F, S, and T, the vowels appear in alphabetic order -- A, E, I, O, U (e.g., as in "FACETIOUS")?"
I ended up with:
My rational being that you order A, E, I, O, U and you have 6 choices of where to put the first of the remaining 4 letters: _ A _ E _ I _ O _ U _ . So, choose one of those letters, and multiply by the 6 possibilities. Once the first letter is placed, it opens up a new spot. Choose the next letter, , and place it in one of the 7 remaining spots. (Same for last two remaining letters).
The marker ended up with:
His rational being that you have 9 spots to fill, choose 5 of them for the vowels, and arrange the remaining four letters in the remaining 4 spots.
I ended up with an additional 4!, but neither of us could see where I (most likely) went wrong.
Any clarification you could provide would be greatly appreciated .
Thanks,
--
Dave
(Hopefully I explained that properly.)
The marker is right.
Suppose at first you chose F and decided to insert it between A and E. Then suppose you chose C and inserted it between F and E. On the other hand, you could choose C first and insert it between A and E, then choose F and insert it between A and C. In both cases, you end up with _ A _F _ C _ E _ I _ O _ U _. Thus, you count several variants twice.