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:

$\displaystyle \begin{align*}P(A) &= \frac{\binom{4}{1} \cdot 6 \cdot \binom{3}{1} \cdot 7 \cdot \binom{2}{1} \cdot 8 \cdot \binom{1}{1} \cdot 9}{9!} \\ &= \frac{(9 \cdot 8 \cdot 7 \cdot 6) \cdot 4!}{9!} \\ &= 0.2 \end{align*}$

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, $\displaystyle \binom{4}{1}$ and multiply by the 6 possibilities. Once the first letter is placed, it opens up a new spot. Choose the next letter, $\displaystyle \binom{3}{1}$, and place it in one of the 7 remaining spots. (Same for last two remaining letters).

The marker ended up with:

$\displaystyle \begin{align*} P(A) &= \frac{\binom{9}{5}4!}{9!} \\ &= \frac{9 \cdot 8 \cdot 7 \cdot 6}{9!} \\ &= 0.0083 \bar{3} \end{align*}$

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.)