Problem 11 - Discrete Mathematics with Applications

Question

Let \(A\) be a 10 -element subset of the set \(\\{1,2, \ldots, 20\\}\) Determine if \(A\) has two five-element subsets that yield the same sum of the elements.

Step-by-Step Solution

Answer

It is guaranteed that there are two five-element subsets of a 10-element subset A of the set {1,2, ..., 20} that yield the same sum. This is demonstrated using the Pigeonhole Principle. There are 76 possible sums for five-element subsets, and there are \(\binom{10}{5} = 252\) five-element subsets of A. Since the number of subsets (252) is greater than the number of possible sums (76), at least one sum must correspond to two different subsets.

Step 1: Pigeonhole Principle

We are going to use the well-known Pigeonhole Principle, which states that if there are n pigeonholes (containers) and m pigeons (objects), and if m > n, then at least one pigeonhole contains more than one pigeon. In this case, the pigeonholes will represent all possible sums of five-element subsets, and the pigeons will correspond to the possible five-element subsets of A.

Step 2: Calculating Number of Possible Sums

First, let's find the range of possible sums for five-element subsets. The minimum sum corresponds to the lowest five elements in the set {1,2, ..., 20}, which are {1, 2, 3, 4, 5}. The sum is given by 1 + 2 + 3 + 4 + 5 = 15 The maximum sum corresponds to the highest five elements in the set {1,2, ..., 20}, which are {16, 17, 18, 19, 20}. The sum is given by 16 + 17 + 18 + 19 + 20 = 90 There are 90 - 15 + 1 = 76 possible sums of five-element subsets.

Step 3: Counting Number of Pigeons

Now, let's calculate the number of pigeons, which are the five-element subsets of the 10-element set A. Using the combination formula, we find there are \(\binom{10}{5}\) five-element subsets of A: \[\binom{10}{5} = \frac{10!}{5!(10 - 5)!} = \frac{10!}{5!5!} = 252\]

Step 4: Applying Pigeonhole Principle

Since we have 76 pigeonholes (possible sums) and 252 pigeons (five-element subsets of A), the number of pigeons is greater than the number of pigeonholes (252 > 76). According to the Pigeonhole Principle, there must be at least one pigeonhole with more than one pigeon, which means that there must exist subsets with the same sum. Therefore, it is guaranteed that there are two five-element subsets of A whose elements sum to the same value.

Key Concepts

Discrete Mathematics

Discrete Mathematics is a branch of mathematics that deals with objects that can assume only distinct, separated values. It plays a crucial role in computer science, cryptography, logic, and combinatorial games, among others. Unlike continuous mathematics that deals with smooth changes, discrete mathematics focuses on countable, often finite sets of elements.

Within Discrete Mathematics, the Pigeonhole Principle is a powerful concept used to prove the existence of certain properties without explicitly identifying them. The principle asserts that if you have more 'pigeons' than 'pigeonholes', then at least two pigeons must share a pigeonhole. This idea is foundational in many areas, including algebra, number theory, and, as in the exercise we’re considering, in problems involving combinations and permutation of sets.

Combination Formula

The Combination Formula is a fundamental concept in combinatorics, a subject within Discrete Mathematics. It allows us to calculate the number of ways a certain number of items can be selected from a larger set without regard for the order of selection.

In mathematical notation, the number of ways to choose k elements from an n-element set is denoted as \(\binom{n}{k} \). This is also referred to as 'n choose k'. The formula is given by:\[\binom{n}{k} = \frac{n!}{k!(n - k)!}\],where '!' denotes factorial, the product of all positive integers up to that number. In the context of our exercise, \(\binom{10}{5} \) calculations demonstrated there is a large number of possible subsets, solidifying the relevance of the Combination Formula in solving such problems.

Subset Sum Problem

The Subset Sum Problem is a classic problem in computer science and Discrete Mathematics, commonly associated with the topic of algorithmic complexity and combinatorics. It involves determining whether a subset of numbers can be found within a larger set that adds up to a given sum.

This problem is broadly applicable, including in budgeting scenarios, to check whether a selection of items can fit into a specified limit. While the Subset Sum Problem can become computationally intensive as set sizes grow, the Pigeonhole Principle offers a clever shortcut to proving the existence of two subsets with equal sums, which is the crux of the problem we've discussed. Our exercise revealed that by understanding the bounds of possible sums and the count of subset combinations, we could apply the Pigeonhole Principle to guarantee at least one pair of five-element subsets from the set \(A\) have identical sums, elegantly circumventing the need for exhaustive enumeration.