Problem 58 - Discrete Mathematics with Applications

Question

Stirling numbers of the second kind \(S(n, r)\) are also given by the formula $$S(n, r)=\frac{1}{r !} \sum_{k=0}^{r-1}(-1)^{k}\left(\begin{array}{l}r \\\k\end{array}\right)(r-k)^{n}$$ Compute each Stirling number. $$S(4,2)$$

Step-by-Step Solution

Answer

The Stirling number of the second kind \(S(4,2)\) is 7.

Step 1: Plug in n and r values

First, we need to plug in \(n=4\) and \(r=2\) in the given formula. We get: \[S(4,2) = \frac{1}{2!} \sum_{k=0}^{1}(-1)^{k}\binom{2}{k}(2-k)^{4}\]

Step 2: Evaluate 2! and simplify

Next, we evaluate the factorial of 2, finding that 2! = 2. Now we simplify the expression: \[S(4,2) = \frac{1}{2} \sum_{k=0}^{1}(-1)^{k}\binom{2}{k}(2-k)^{4}\]

Step 3: Calculate terms of the sum

Now we will calculate the terms of the sum by plugging in \(k=0\) and \(k=1\) into the summand. When \(k=0\): \((-1)^{0}\binom{2}{0}(2-0)^{4} = 1 \cdot 1 \cdot 16 = 16\) When \(k=1\): \((-1)^{1}\binom{2}{1}(2-1)^{4} = -1 \cdot 2 \cdot 1 = -2\)

Step 4: Sum the terms and multiply by 1/2

Now we need to sum the terms and multiply the result by 1/2. We get: \[S(4,2) = \frac{1}{2} (16 - 2) = \frac{1}{2} \cdot 14 = 7\] Thus, the Stirling number of the second kind \(S(4,2)\) is 7.

Key Concepts

Combinatorics

Combinatorics is a fascinating area of mathematics that deals with counting, arrangement, and combination of objects. It provides us with the tools to solve problems related to discrete structures. At its core, combinatorics helps us answer the question: "In how many ways can a certain event occur?" This is essential for understanding various scenarios, from organizing books on a shelf to determining possible outcomes of complex systems.

• **Applications**: It plays a crucial role in computer science, cryptography, and even in the arrangement of seating at events.
• **Key Concepts**: Permutations (arrangements) and combinations (selections) are fundamental to understanding combinatorics.

Stirling numbers of the second kind, which are part of combinatorial mathematics, help us determine the number of ways to partition a set of objects into non-empty subsets. This is particularly useful in scenarios where grouping is required, such as distributing tasks among team members.

Factorials

Factorials are mathematical operations that involve multiplying a series of descending natural numbers. Denoted by an exclamation mark, for a positive integer \( n \), the factorial is expressed as \( n! \).

• For example, \( 4! = 4 \times 3 \times 2 \times 1 = 24 \).
• By definition, \( 0! = 1 \), which is a convention used to ensure that the formulae involving factorials are consistent.

In the context of Stirling numbers, factorials are used to effectively distribute elements. When we compute Stirling numbers like \( S(4,2) \), we use a factorial in the denominator to normalize the number of ways items can be grouped into subsets. This ensures each partitioning is considered uniquely, accounting for the arrangements of objects in subsets.

Binomial Coefficients

Binomial coefficients, often written as \( \binom{n}{k} \), represent the number of ways to choose \( k \) objects from a set of \( n \) objects without regard to the order of selection. They play a significant role in algebra, particularly in the expansion of expressions raised to a power, known as the binomial theorem.

• The formula for computing binomial coefficients is \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \).
• They appear in various areas, such as statistical calculations and probability theory.

In relation to Stirling numbers, binomial coefficients help compute the sum in the Stirling formula. They weigh each term in the expansion, modifying the count of partitions to match particular conditions—ensuring precise partitioning in accordance with each subset's arrangement. While solving for \( S(4,2) \), the coefficient \( \binom{2}{k} \) aids in determining the contribution of each possible subset size to the overall number of partitions.