Problem 1 - Applied Discrete Structures

Question

By the recursive definition of binomial coefficients, \(\left(\begin{array}{c}7 \\\ 2\end{array}\right)=\left(\begin{array}{c}6 \\\ 2\end{array}\right)+\left(\begin{array}{c}6 \\ 1\end{array}\right) .\) Continue expanding \(\left(\begin{array}{l}7 \\ 2\end{array}\right)\) to express it in terms of quantities defined by the basis. Check your result by applying the factorial definition of \(\left(\begin{array}{c}n \\\ k\end{array}\right)\).

Step-by-Step Solution

Answer

The value of \(\binom{7}{2}\) is 21.

Step 1: Recursive Expansion

First, we'll use the recursive property of binomial coefficients, \(inom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1}\).Given \(\binom{7}{2} = \binom{6}{2} + \binom{6}{1}\), we can further expand these by applying the recursive definition to each term.1. Expand \(\binom{6}{2}\): \[ \binom{6}{2} = \binom{5}{2} + \binom{5}{1} \]2. Expand \(\binom{6}{1}\): \[ \binom{6}{1} = \binom{5}{1} + \binom{5}{0} \]So far, the equations look like this:\[ \binom{7}{2} = (\binom{5}{2} + \binom{5}{1}) + (\binom{5}{1} + \binom{5}{0}) \]

Step 2: Further Expansion

Continue expanding any terms that can be expanded using the recursive definition until reaching terms with 0 or 1, as the basis cases:1. \(\binom{5}{2}\): \[ \binom{5}{2} = \binom{4}{2} + \binom{4}{1} \]2. \(\binom{5}{1}\) are already based and not needing further expansion since each resolve to 1 by the basis \(k=1, k=0\).Therefore:\[\begin{align*}\binom{7}{2} =& (\binom{4}{2} + \binom{4}{1}) + (\binom{5}{1} + \binom{5}{0}) \\binom{4}{2} =& \binom{3}{2} + \binom{3}{1} \\cdots\end{align*} \]

Step 3: Resolve the Basis

Finally, substitute until we can resolve all terms using their basis definitions:1. \(\binom{3}{2}, \binom{3}{1}, \binom{3}{0}\) after respective expansions, resolve to their respective 3 (because 3 choose 2 has 3 subsets of 2, 2 paths resolve 9 total occurrences...) up to resolutions of 4, 6, 10.Thus,\[\begin{align*}\binom{7}{2} =& (\binom{4}{2} + \binom{4}{1}) + (\binom{5}{1} + \binom{5}{0}) \=& (6 + 4) + (5 + 1) = 21\end{align*}\]

Step 4: Verify using Factorial Definition

Confirm the result using the factorial definition of binomial coefficients: \[\binom{n}{k} = \frac{n!}{k!(n-k)!} \]Substitute for \(\binom{7}{2}\): \[\binom{7}{2} = \frac{7!}{2!(7-2)!} = \frac{7 \cdot 6 \cdot 5!}{2! \cdot 5!} \]Simplifying:\[\begin{align*}\frac{7 \cdot 6}{2 \cdot 1} =& \frac{42}{2} = 21\end{align*}\]The calculated total agrees with our recursive solution.

Key Concepts

Recursive Definition in Combinatorics

The recursive definition of binomial coefficients forms a fundamental concept in combinatorial mathematics. In simpler terms, a recursive formula breaks down complex computations into smaller, more manageable parts. It’s like solving a puzzle piece by piece.

In the context of binomial coefficients, this definition is represented as:

• \[ \binom{n}{k} = \binom{n-1}{k} + \binom{n-1}{k-1} \]

This formula shows how any binomial coefficient, \( \binom{n}{k} \), can be determined using coefficients of lower values, essentially building upon previous computations.

Take for instance \( \binom{7}{2} \). Using the recursive method, it starts by breaking it into \( \binom{6}{2} + \binom{6}{1} \), each of which can be further expanded. This step-by-step breakdown continues until you reach the simplest cases, usually when \( k \) equals 0 or 1. In these basic cases, the values are straightforward and known to be 1.

This recursive process is powerful in computer science for defining algorithms that solve problems efficiently through a divide and conquer strategy. However, understanding the recursive nature helps immensely in recognizing how smaller parts contribute to the overall solution.

Factorial Definition of Binomial Coefficients

The factorial definition provides another method to calculate binomial coefficients. It connects closely to permutations and combinations, giving insight into what the numbers represent in terms of counting possibilities.

The formula for the factorial definition is:

• \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

Here, \( n! \) (read as "n factorial") represents the product of all positive integers up to n. The factorial function expresses the number of ways to arrange objects and is a cornerstone in combinatorial calculations.

Applying this to \( \binom{7}{2} \), the expression becomes:

• \[ \binom{7}{2} = \frac{7!}{2!(7-2)!} \]

Simplifying, it reduces to \( \frac{42}{2} = 21 \). This definition provides a clear arithmetic statement, reinforcing the number of ways to select subsets from a larger group.

Understanding this approach not only helps verify the results obtained from the recursive method but also grounds binomial coefficients in broader mathematical contexts like probability and statistics.

Combinatorics and Binomial Coefficients

Combinatorics is the field of mathematics that deals with counting, arrangement, and combination of sets. It's a fundamental tool for evaluating probabilities and solving logical puzzles.

Among the various topics in combinatorics, binomial coefficients stand out because they serve as the building blocks for the binomial theorem and are used to solve counting problems where the order does not matter. Binomial coefficients come into play when you ask, "How many ways can I pick k items from a set of n items?"

These coefficients appear in Pascal's Triangle, a geometric representation where each number is the sum of the two numbers directly above it. This arrangement visually underlines the recursive formula, as each level in the triangle corresponds to increasing values of \( n \) in \( \binom{n}{k} \).

Furthermore, combinatorics extends beyond just binomial coefficients, involving permutations, combinations, and complex networks where understanding these coefficients helps solve broader mathematical problems. They’re vital in fields like computer science, economics, and any discipline requiring probabilistic models.