In the usual classification, L is a consonant. On the assumption it is a vowel, the cases division that you made is correct. The approach could be modified to deal with the fact that L is a consonant. It would become somewhat more complicated.
We solve the problem with L a consonant, and without dividing into cases. Put ID's on the letters that are the same, to make them distinct. So we have $9$ different letters, of which $5$ are consonants.
The first slot can be filled with a consonant in $5$ ways. For each such way, the last slot can be filled in $4$ ways. And then the "middle" can be filled in $7!$ ways, for a total of $(5)(4)(7!)$.
Now take off the ID numbers. When we take off the ID's on the A's, the number of distinct words gets divided by $2!$. There is also division by $2!$ when we remove the ID's from the C's, and from the L's. We end up with
$$\frac{(5)(4)(7!)}{(2!)^3}$$
distinct words that satisfy the specification.
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
One say to think of it is to write the letters C and S like this: C C S S S
The letter E can be inserted into any of the $6$ "gaps" (these include the $2$ endgaps). And for every way to insert the E, there are $7$ places to insert the U.