Thread: Permutation of items with a constant order subset.

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.)

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

The marker is right.

Originally Posted by driegert

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).

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.

Thanks! I was thinking I had double permuted the remaining letters.

Originally Posted by driegert

"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")?"

There is a simple way to look at this problem.
The answer is simply

Look at it this way. Only 1 in every 120 arrangements do the vowels appear in alphabetic order.
It is the same answer if the set were
.