Solution: This is equivalent to partitioning 6 distinct compounds into 3 non-empty, unordered subsets (phases). The number of ways is the Stirling number of the second kind $ S(6, 3) $, multiplied by the number of ways to assign these subsets to 3 distinguishable phases (since phases are distinct, e.g., Phase 1, 2, 3). - iBuildNew

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