kapslock wrote:

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!

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