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 \(\fbox{4760}\)
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