So, the average length is $ \boxed{6} $ centimeters.Question: An archaeologist is analyzing 7 distinct artifacts found at an Inca settlement, and wishes to divide them into 3 identical storage containers such that each container holds at least one artifact. How many distinct ways can this be done?

["So, the average length is $ \boxed{6} $ centimeters? No—today’s insightful exploration reveals a combinatorial puzzle far richer than a simple average.", "When an archaeologist uncovers 7 distinct artifacts from an Inca settlement, a key logistical question arises: How many distinct ways can these 7 unique items be divided into 3 identical storage containers, with each container holding at least one artifact? This problem lies at the heart of combinatorics and wants simple yet precise solutions.", "### Understanding the Problem", "We are not just packing objects—we are partitioning a set of 7 distinct artifacts into 3 non-empty, unlabeled (identical) containers. This rules out standard permutation or combination formulas, and demands careful use of Stirling numbers of the second kind.", "Let’s define:", "- $ S(n, k) $: the Stirling number of the second kind, representing the number of ways to partition a set of $ n $ distinct objects into $ k $ non-empty, unlabeled subsets.\nHere, $ n = 7 $, $ k = 3 $. We seek $ S(7, 3) $.", "### Calculating $ S(7, 3) $", "There are recursive formulas to compute Stirling numbers, but a known value is:", "$$\nS(7, 3) = 301\n$$", "This number counts all distinct ways to partition 7 labeled objects into 3 non-empty unlabeled groups—exactly the scenario the archaeologist faces.", "However, the question specifies that containers hold distinct artifacts (so labeling matters in content, but the groups are indistinct in order). Since the containers are identical, arrangements like {A} in box 1 and {B,C,D} in box 2 vs. {B,C,D} in box 1 and {A} in box 2 are indistinguishable—hence the use of $ S(7,3) $, not multiplied by permutations.", "Additionally, the constraint that each container holds at least one artifact excludes solutions with fewer than 3 non-empty containers. But $ S(7,3) $ already enforces this: no empty subsets.", "### Final Answer", "Thus, the number of distinct ways the archaeologist can divide the 7 distinct artifacts into 3 identical storage containers, each holding at least one artifact, is:", "$$\n\boxed{301}\n$$", "---", "Bonus Insight:\nThis reflects a broader principle: when grouping unique items into unlabeled, non-empty containers, Stirling numbers of the second kind provide an elegant, accurate combinatorial foundation—ideal for archaeological logistics, inventory systems, and more.", "Length note (as hinted): The value 301 emerges naturally from recursion:\n$ S(n,k) = k \cdot S(n-1,k) + S(n-1,k-1) $,\nwith base cases. While full derivation exceeds brevity, this number appears prominently in standardized combinatorics—confirming its reliability.", "So yes—though the average length mentioned in the title is a modest $ \boxed{6} $ cm, the real treasure lies in solving the deeper puzzle of how many meaningful, balanced groupings exist—a number as deliberate as it is profound."]