Now compute $S(5,2) = 2 \cdot S(4,2) + S(4,1) = 2 \cdot 7 + 1 = 14 + 1 = 15$
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- Solution: We are partitioning $6$ distinguishable objects (proposals) into $3$ non-empty, indistinct subsets (teams). This is given by the Stirling number of the second kind $S(6,3)$.
- We compute $S(6,3)$ using recurrence:
- $S(6,3) = 3 \cdot S(5,3) + S(5,2)$
- S(6,3) = 3 \cdot 25 + 15 = 75 + 15 = 90
- Thus, the number of ways is $\boxed{90}$.**Question:
- Find the length of the shortest altitude in a triangle with sides measuring 13, 14, and 15 units.
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