Your first solution ($36000$) counts words in which AEU don't come all three together, your second solution ($14400$) counts those in which they are all three separate, which is quite a more strict condition. Which shows that it helps to pose your question more carefully: what exactly do you mean by "the vowels come together"?
I am not very sure about the methodology. Please let me know if the logic is faulty anywhere.
We have $3$ slots.$$---$$
The middle one has to be a vowel. There are only two ways in which it can be filled: O, A.
Let us put O in the middle.$$-\rm O-$$
Consider the remaining letters: O, A, $\bf P_1$, $\bf P_2$, R, S, L. Since we will be getting repeated words when we use $\bf P_1$ or $\bf P_2$ once in the word, we consider both these letters as a single letter P. So, the letters now are O, A, P, R, S, L. We have to select any two of them (this can be done in $^6C_2$ ways) and arrange them (this can be done in $2!$ ways). So, total such words formed are $(^6C_2)(2!)=30$. Now there will be one word (P O P) where we will need both the P's. So we add that word. Total words=$31$.
Now, we put A in between.$$-\rm A-$$
Again proceeding as above, we have the letters O, P, R, S, L. Total words formed from them are $(^5C_2)(2!)=20$. We add two words (O A O) and (P A P). Thus, total words formed here=$22$
Hence, in all, total words that can be formed are $22+31=53$.
(P.S: Please edit if anything wrong.)
Best Answer
All the letters are different, so that makes things easier.
Pick the two vowels ($_3C_2$) and pick the three consonants ($_5C_3$) and then pick what order they go in $(5!)$. So the answer is $3 \cdot 10 \cdot 120 = 3600.$
You take combinations of the vowels and consonants because the order of them doesn't matter at that point. You order them in the last step, after you've chosen which ones go in your five-letter word.
In other words, it doesn't matter that I pick $A$, then $U$, instead of $U$, then $A$. It just matters that I picked the set $(A,U)$.