There are in total

\(\frac{8!}{2!2!}\) total arrangements of letters

considering arrangements of letter where the two 'L's are right next to each other, we have

\(\frac{7!}{2!}\) total arrangements

and we have the same amounts of arrangements for the two 'S's being rigth next to each other

So therefore, we have

\(\frac{8!}{2!2!} - 2 * \frac{7!}{2!} + 6!\) because we overcount a bit

so the answer is \(\fbox{4760}\)

.