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
The problem is that you assume that each vowel and consonant can only occur one time. However, some vowels and consonants appear multiple times. One way to solve this problem is the following. First, let us select three vowels and their respective order:
This results in a total of 43 possible combinations. Second, let us select four consonants and their respective order:
This results in a total of 354 possible combinations. Selecting three positions to place the vowels, the total number of words thus equals:
$$43 \cdot 354 \cdot {7 \choose 3} = 532770$$