$\{1,2,3\}, \{4\}$

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• $S(4,2) = 2^2 - 1 = 7$? No — standard recurrence:
• $S(3,2) = 2 \cdot 1 + 1 = 3$
• Alternatively: list all such partitions:
• $\{1,2,4\}, \{3\}$
• $\{1,3,4\}, \{2\}$
• $\{2,3,4\}, \{1\}$

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