S(7,3) = 3 \cdot 90 + 31 = 270 + 31 = 301.

February 22, 2026

$S(7,3) = 3 \cdot 90 + 31 = 270 + 31 = 301.$

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$ S(6,3) = 90 $
$ S(7,3) = 3 \cdot S(6,3) + S(6,2) $
We compute $ S(6,2) = 2 \cdot S(5,2) + S(5,1) = 2 \cdot 15 + 1 = 31 $, or known $ S(6,2) = 31 $. Then:
Thus, the number of ways to partition the 7 distinct artifacts into 3 indistinguishable non-empty containers is $ \boxed{301} $.
Question: A science educator is designing a virtual lab where students simulate flipping 6 fair coins and rolling a single 6-sided die. What is the probability that the number of heads equals the value rolled on the die?
Solution: Let $ H $ be the number of heads in 6 flips of a fair coin, so $ H \sim \text{Binomial}(6, \frac{1}{2}) $. Let $ D $ be the outcome of a fair 6-sided die, uniformly distributed over $ \{1,2,3,4,5,6\} $. We seek $ P(H = D) $.

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