Permutations of words from original word. How many distinct 4-letter arrangements can be made with the letters in the word "PARALLEL"?
My approach:
Because we are only looking at how many different permutations there are and not the frequency at which these permutations exist, we can delete the repeated letters and leave only one. This leaves us with the following set of letters: $\{P, A, R, L, E\}$ So $5$ permute $4$ is $120$. It was only then that I realized that deleting repeats will remove words such as $LLLE$  to exist. At this point, I do not know how to add on these possibilities to my approach. 
I would appreciate help, and as always, bash me whenever you see a typical blunder of mine.

There are a few cases:


*

*All $4$ letters distinct. There are $\binom{5}{4} = 5$ ways to pick the letters and $4!$ ways to order them.

*Two letters distinct, one letter repeated twice. Either two A's or two L's. If we have two A's, then we have $\binom{4}{2} = 6$ ways to pick the other letters. $4!$ ways to order them, but divide by $2!$ to account for the doubled letter. Then do the same thing again for the double-L case.

*Two letters repeated twice. Two A's, two L's. $4!$ ways to order the letters, divide by $2!2!$ to account for the repetitions.

*One letter by itself, one letter repeated three times. Just the L here. Once we have the triple L, there are $4$ choices for the final letter. $4!$ ways to arrange the letters, divided by $3!$ for the triple L.
$$5 \cdot 4! + \frac{6 \cdot 4! \cdot 2}{2!} + \frac{4!}{2! \cdot 2!} + \frac{4 \cdot 4!}{3!} = 286$$
