That's what I'd go for too.
5!/2 - 4!.
This is easy since the question read at least one space (alphabet) between the two Ls.
Try the variation of this question of finding the permutations with exactly 1 space (alphabet) between the two Ls.
Wud post my answer in a while.
Two ways of approaching this problem:
1. Consider L X L as a group, where X can be O, C or A.
Thus the variations are (L X L) Y Y where Y Y are the remaining characters.
For one value of X, the possible permutations are 3!. (Two Ys and one LXL).
For all possible three values of X, the possible permutations = 3 * 3! = 18
2. Consider the positions
L _ L _ _ - 3 positions, 3! variations possible
_ L _ L _ - 3 positions, 3! variations possible
_ _ L _ L - 3 positions, 3! variations possible
Total = 3 * 3!
Who says elephants can't dance?