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.

If you use $j$ A's and $k$ L's, there are $\binom3{4-j-k}$ ways to select the remaining letters, and there are $\frac{4!}{j!k!}$ different ways to arrange the $4$ letters, so in total there are
$$
\sum_{j=0}^2\sum_{k=0}^3\binom3{4-j-k}\frac{4!}{j!k!}=286
$$
admissible words (Wolfram|Alpha computation of the double sum).
