In how many ways can you arrange the word "director" so that the vowels' orders stay the same? 
In how many ways can you arrange the word "director" so that the vowels' orders stay the same?

I found an answer where it said we find all possible permutations of this word and then divide it by the possible permutations of the vowels. The possible permutation of the vowels i, e, o are $3! = 6$.
The $6$ possible arrangements of the vowels are: ieo (the same as in the word), ioe, eio, oie, eoi, oei.
Now, we're supposed to keep the arrangements with the order ieo, then why do we have to divide with $6$ instead of $5$? What am I missing here?
 A: First imagine listing off all arrangements of "director".  Write each down on a piece of paper. Now, on the back of each piece of paper, replace each vowel with an 'x'. eg, director -> dxrxctxr, etc.  Now put all the pieces of paper into folders, with each word going into a folder labeled by the "x-ed" version of that word.
dxrxtxr
director dirocter derictor deroctir dorectir doricter
drxxtxr
driector driocter dreictor dreoctir droectir droicter
etc.
You'll find that in each folder, there are exactly 6 words, exactly one of which has the vowels in the order ieo, so you'll get one word matching your condition for each folder. So, the answer is given by taking the total number of arrangements and dividing by 6. Now, you just need to count the total number of arrangements of director. (Careful - remember there are two rs!)
A: DIRECTOR has vowels $IEO$. There are $6$ possible orders of vowels. The question to ask is this - will one of the orders of vowels have more number of words than others (for the given letters)? If you are still not convinced, take a simpler example -
$AIDE$
Total number of permutations of letters is $4! = 24$
But with the order of vowels fixed, there are $4$ possible words as $D$ can be in $4$ places -
$ \uparrow A \uparrow I \uparrow E \uparrow$
For every possible order of vowels, there are $4$ possible words. That leads to total number of words as,
$3! \cdot 4 = 4! = 24$
There is an alternate solution -
Place $IEO$ first in that order. Now place two $R's$ for which you can choose two places out of four or you can choose one of the four places. That is $10$ ways in total. Now you have $5$ letters and that gives choice of six places for the next letter, similarly seven for the next and eight for the last.
That is $10 \cdot 6 \cdot 7 \cdot 8 = 3360$
A: Assume first that we distinguish all the letters, particularly both 'R's (like one is red and one is blue).
Imagine somebody made $8!$ arrangements. You found that in many cases vowels are in wrong order. You made corrections and realised that each arrangement is present six times in your list of obtained arrangements. Therefore you have to divide by six.
You can solve the task in other way. First, you choose the three places for vowels, that is $\binom 83$. These places define precise position of the vowels. Now you can arrange 5 consonants in $5!$ ways. Therefore the number of arrangements is $\binom 83\cdot 5!=8!/3!$, which matches the previous result.
Now we can deal with two Rs just like with vowels. Since interchanging their position doesn't change the arrangement, we have to divide the number of arrangements by two.
A: If there are 3 vowels, then there are 3! (or 6) possible ways to permute them. Therefore, 1/6 of the arrangements have the vowel arrangement, “ieo,” for any word including these vowels.
You would divide the permutation of letters in “director” by 6 (or multiply the permutation by 1/6) to get rid of the cases where the vowels are not arranged as “ieo.” After that, deal with the two Rs by dividing by 2! (or 2).
If you were to divide by 5 instead, that would suggest that 1/5 of the permutations have the favorable vowel arrangement, “ieo,” which is false. Dividing by the number of unfavorable permutations for the three vowels would not give the right answer, as you would need to divide by every permutation of the vowels to get the permutations of “director” that follow that one specific vowel arrangement.
Therefore, the answer should be 8!/12.
