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๐ŸงฎCombinatorics Unit 8 Review

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8.2 Stirling numbers of the first kind

๐ŸงฎCombinatorics
Unit 8 Review

8.2 Stirling numbers of the first kind

Written by the Fiveable Content Team โ€ข Last updated September 2025
Written by the Fiveable Content Team โ€ข Last updated September 2025
๐ŸงฎCombinatorics
Unit & Topic Study Guides

Stirling numbers of the first kind are crucial in counting permutations with specific cycle structures. They're like the secret code to understanding how elements can be arranged in different cycles, helping us solve complex counting problems.

These numbers connect to the broader world of partitions by showing how elements can be grouped. They're a key tool in our combinatorial toolkit, letting us tackle tricky permutation puzzles and unlock deeper insights into mathematical patterns.

Stirling Numbers of the First Kind

Definition and Combinatorial Interpretation

  • Stirling numbers of the first kind s(n,k) count permutations of n elements with exactly k disjoint cycles
  • Unsigned s(n,k) and signed [n k] versions exist with [n k] = (-1)^(n-k) s(n,k)
  • s(n,k) represents ways to arrange n distinct objects into k nonempty cycles
  • Related to falling factorial powers x(n)=x(xโˆ’1)(xโˆ’2)...(xโˆ’n+1)=โˆ‘k=0ns(n,k)xkx^{(n)} = x(x-1)(x-2)...(x-n+1) = \sum_{k=0}^n s(n,k) x^k
  • Sum of absolute values for fixed n equals n! (total permutations of n elements)
  • Represented in triangular array similar to Pascal's triangle
    • Differs in recurrence relations used to generate values

Properties and Relationships

  • Directly connected to cycle structure of permutations in symmetric group Sn
  • s(n,k) gives number of n-element permutations with k cycles
  • Sum of s(n,k) for fixed n equals n! โˆ‘k=1ns(n,k)=n!\sum_{k=1}^n s(n,k) = n!
  • Used to count permutations with specific cycle structures (fixed points)
  • Exponential generating function related to logarithm of permutation generating function
  • Connected to coefficients in rising factorial expansion (x)n=x(x+1)(x+2)...(x+nโˆ’1)(x)_n = x(x+1)(x+2)...(x+n-1)

Recurrence Relations for Stirling Numbers

Fundamental Recurrence Relation

  • Main recurrence relation s(n+1,k)=s(n,kโˆ’1)+ns(n,k)s(n+1,k) = s(n,k-1) + ns(n,k)
  • Combinatorial interpretation breaks down into two cases
    • Add new element as single cycle to n elements in k-1 cycles
    • Insert new element into n positions in k cycles of n elements
  • Initial conditions
    • s(n,0) = 0 for n > 0
    • s(0,0) = 1
    • s(n,k) = 0 for k > n
  • Enables efficient computation for large n and k values

Alternative Recurrence Relations

  • Another useful relation s(n+1,k)=โˆ‘i=kn(ni)s(i,k)s(n+1,k) = \sum_{i=k}^n \binom{n}{i} s(i,k)
    • Applicable in certain problem-solving scenarios
  • Recurrence for signed Stirling numbers [n+1ย k]=[nย kโˆ’1]โˆ’n[nย k][n+1 \space k] = [n \space k-1] - n[n \space k]
  • Various recurrence relations offer flexibility in different contexts

Stirling Numbers and Permutations

Cycle Structure Analysis

  • s(n,k) counts n-element permutations with k cycles
  • Sum of s(n,k) for fixed n equals total permutations (n!)
  • Used to analyze permutations with specific cycle structures
    • Example: Counting permutations with 2 fixed points
  • Helpful in studying random permutations and their cycle structures

Generating Functions

  • Exponential generating function for s(n,k) linked to permutation generating function logarithm
  • Useful for solving complex enumeration problems
  • Provides insights into relationships between permutations and Stirling numbers

Problems with Stirling Numbers

Computation and Counting

  • Use recurrence relations to calculate s(n,k) for given n and k
    • Example: Compute s(5,3) using the fundamental recurrence relation
  • Apply combinatorial interpretation to solve cycle arrangement problems
    • Example: Ways to arrange 6 people into 2 circular tables
  • Analyze permutations with specific cycle structures
    • Example: Count 7-element permutations with exactly 3 cycles

Advanced Applications

  • Employ generating functions for complex enumeration
    • Example: Find the number of permutations of 8 elements with no fixed points
  • Use relationship with falling factorial powers in algebraic manipulations
    • Example: Expand x^(5) in terms of Stirling numbers
  • Solve problems involving both first and second kind Stirling numbers
    • Example: Relate s(n,k) to S(n,k) (Stirling numbers of the second kind)
  • Apply in probability theory for random permutation analysis
    • Example: Probability of a random 9-element permutation having exactly 4 cycles