Total number of arrangements of the letters of the word SUCCESS such that both $C's$ are together and no two $S's$ are together is

Treat the two C's as a single block, leaving 6 items to arrange: U, E, the CC block, and three S's. The number of ways to arrange these 6 items without any restrictions is 6! / 3! = 120 ways. To ensure no two S's are together, we must use the gap method: place the 3 non-S items first, creating 4 gaps (including ends), and arrange the 3 S's in these gaps, yielding 120 / 3! * 4C3 = 24 valid arrangements.

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