Solution: This is the Stirling number of the second kind $ S(6, 3) $, calculated as $ \frac{1}{3!} \left( 3^6 - 3 \cdot 2^6 + 3 \cdot 1^6 \right) = \frac{729 - 192 + 3}{6} = \frac{540}{6} = 90 $.

["Understanding the Stirling Number of the Second Kind: Solving ( S(6, 3) = 90 )", "In combinatorics and discrete mathematics, Stirling numbers of the second kind, denoted ( S(n, k) ), play a crucial role in counting the number of ways to partition a set of ( n ) distinct elements into ( k ) non-empty, unlabeled subsets. These numbers are not just abstract constructs—they appear in diverse fields such as probability, algebra, and computer science. One commonly studied value is the Stirling number ( S(6, 3) ), which equals 90. In this article, we explore what ( S(6, 3) ) represents, derive its value using the formula ( S(6, 3) = \frac{1}{3!} \left( 3^6 - 3 \cdot 2^6 + 3 \cdot 1^6 \right) ), and understand its significance.", "---", "### What Is the Stirling Number of the Second Kind?", "The Stirling number of the second kind, ( S(n, k) ), counts the number of ways to partition a set of ( n ) labeled objects into exactly ( k ) non-empty, unlabeled subsets. For example, ( S(4, 2) ) tells us how many ways we can divide 4 distinct people into 2 unlabeled groups—such as team pairings—where each group has at least one person.", "This concept is essential in problems involving grouping, clustering, or distribution—especially when the order of groups does not matter.", "---", "### The Formula for ( S(n, k) )", "The general formula for computing ( S(n, k) ) is:", "[\nS(n, k) = \frac{1}{k!} \sum_{i=0}^{k} (-1)^{k-i} \binom{k}{i} i^n\n]", "Alternatively, using inclusion-exclusion, a well-known expression is:", "[\nS(n, k) = \frac{1}{k!} \sum_{i=0}^{k} (-1)^{k-i} \binom{k}{i} i^n\n]", "For ( n = 6 ) and ( k = 3 ), this becomes:", "[\nS(6, 3) = \frac{1}{3!} \left( 3^6 - 3 \cdot 2^6 + 3 \cdot 1^6 \right)\n]", "This formula computes the number of partitions of 6 labeled items into 3 unlabeled non-empty subsets.", "---", "### Step-by-Step Calculation: ( S(6, 3) = 90 )", "1. Compute powers:\n ( 3^6 = 729 )\n ( 2^6 = 64 ) → multiplied by 3: ( 3 \cdot 64 = 192 )\n ( 1^6 = 1 ) → multiplied by 3: ( 3 \cdot 1 = 3 )", "2. Apply signs via inclusion-exclusion:", "[\nS(6, 3) = \frac{1}{6} \left( 729 - 192 + 3 \right) = \frac{1}{6} (540) = 90\n]", "Hence, there are 90 distinct ways to partition a set of 6 elements into 3 unlabeled, non-empty subsets.", "---", "### Why ( S(6, 3) = 90 ) Matters", "This value surfaces in various real-world and theoretical contexts:", "- Partitioning data: When organizing 6 distinct records into 3 groups—such as clustering users by behavior—( S(6, 3) ) counts valid groupings without regard to order.\n- Combinatorics problems: It helps solve complex counting problems involving set partitions.\n- Probability and statistics: Used in modeling equal-sized sample partitions and random distributions.\n- Algorithm design: Applied in dynamic programming and graph partitioning algorithms.", "Understanding and computing Stirling numbers like ( S(6, 3) ) deepen insight into discrete structure and openness around the concept of grouping and symmetry in sets.", "---", "### Final Thoughts", "The Stirling number of the second kind ( S(6, 3) = 90 ) elegantly captures the number of ways to divide 6 distinguishable items into 3 non-empty indistinct subsets. Through its precise formula and clear derivation, ( S(6, 3) = \frac{1}{6}(3^6 - 3 \cdot 2^6 + 3 \cdot 1^6) = 90 ), this value exemplifies the powerful intersection of algebra and combinatorial logic. Whether you're solving mathematical puzzles, analyzing algorithms, or modeling real systems, Stirling numbers provide essential tools for thinking in terms of partitions and connectivity.", "---", "Keywords: Stirling number of the second kind, ( S(6, 3) ), combinatorics, partitioning, combinatorial mathematics, set partitioning, inclusion-exclusion principle, ( S(n, k) ), math tutorial, discrete mathematics."]