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Chapter 6: Problem 8

Eighteen persons have first names Alfie, Ben, and Cissi and last names Dumont and Elm. Show that at least three persons have the same first and last names.

Short Answer

Expert verified
Using the Pigeonhole Principle, it is proven that at least three persons have the same first and last names.

Step by step solution

01

Understand the scenario

The problem provides 3 possible first names (Alfie, Ben, and Cissi) and 2 possible last names (Dumont and Elm). If you multiply these, there are \(3 \times 2 = 6\) possible combinations for names. There are 18 persons which need to be distributed among these 6 possible combinations. As per the Pigeonhole Principle, if the number of items is more than the number of containers, at least one container must contain more than one item.
02

Apply the Pigeonhole Principle

To find the minimal number of items in a container using the Pigeonhole principle, divide the total number of items by the number of containers and round up to the next integer. In this case, the calculation would be \( \lceil \frac{18}{6} \rceil = 3 \). Where \lceil \rceil denotes the ceiling function, which rounds up to the next integer.
03

Conclude

Since in the worst case scenario, two people sharing the same first and last names from the 18 available, then using the Principle, at least three people must have the same first and last names. Hence, it's proven by the Pigeonhole Principle.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Combinatorics
Combinatorics is a fascinating area of mathematics that deals with counting, arrangement, and combination of objects. In the given problem, we explore the possible combinations of first and last names for a group of people. With 3 first names and 2 last names, we calculate the total combinations by multiplying the two sets:
  • 3 first names: Alfie, Ben, Cissi
  • 2 last names: Dumont, Elm
Therefore, the total number of combinations is calculated as follows: \[3 \times 2 = 6\]This means we have 6 unique full name combinations. Understanding these combinations gives us a clear picture of how the different pieces (first and last names) come together in the puzzle of this combinatorial problem.
Combinatorics serves as a foundation for many problems in discrete mathematics and is crucial for devising strategies in problem-solving.
Discrete Mathematics
Discrete mathematics focuses on the study of mathematical structures that are fundamentally countable and separate rather than continuous. It's a branch of mathematics dealing with discrete elements that use distinct values. The given problem is set in the realm of discrete mathematics because you are dealing with distinct, separate name combinations.
  • Consider each name combination as a distinct "bucket" or "container."
  • Each person from the 18 must be placed into one of these buckets.
This is where the Pigeonhole Principle鈥攁 concept within discrete mathematics鈥攃omes in handy. The principle helps us determine how items (people in this case) must be grouped based on the given limitations (name combinations), thus helping us see that some "buckets" will inevitably contain more than the initially assumed number of people.
Problem-Solving Strategy
When tackling problems like these, it's crucial to use logical and well-tested strategies. The Pigeonhole Principle is a classic problem-solving strategy that dictates if you have more items than containers, at least one container must hold more than one item. This straightforward principle is applied to solve the presented problem effectively.Here's how you leverage it:
  • Calculate the number of available "containers." In this problem, the containers are the 6 possible name combinations.
  • Distribute the 18 people among these containers.
  • Since 18 is greater than 6, using the ceiling function, we find that at least one combination holds:\[\lceil \frac{18}{6} \rceil = 3\]
This strategy helps prove that at least three people must share the same first and last names. By using such structured approaches, one is able to break down complex problems into manageable parts and provide a clear and definitive solution.

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Most popular questions from this chapter

Use mathematical induction to show that if \(E_{1}, E_{2}, \ldots, E_{n}\) are events, then $$ P\left(E_{1} \cup E_{2} \cup \ldots \cup E_{n}\right) \leq \sum_{i=1}^{n} P\left(E_{i}\right) $$

Find the probability that among \(n \geq 3\) persons, at least three people have birthdays on the same month and date (but not necessarily in the same year). Assume that all months and dates are equally likely, and ignore February 29 birthdays.

Answer to give an argument that shows that in a group of 10 persons there are at least two such that either the difference or sum of their ages is divisible by \(16 .\) Assume that the ages are given as whole numbers. Let \(a_{1}, \ldots, a_{10}\) denote the ages. Let \(r_{i}=a_{i} \bmod 16\) and let $$ s_{i}=\left\\{\begin{array}{ll} r_{i} & \text { if } r_{i} \leq 8 \\ 16-r_{i} & \text { if } r_{i}>8 \end{array}\right. $$ Explain why \(s_{j}=s_{k}\) for some \(j \neq k\).

What is wrong with the following argument, which supposedly counts the number of partitions of a 10 -element set into eight (nonempty) subsets? List the elements of the set with blanks between them: $$x_{1}-x_{2}-x_{3}-x_{4}-x_{5}-x_{6}-x_{7}-x_{8}-x_{9}-x_{10}$$ Every time we fill seven of the nine blanks with seven vertical bars, we obtain a partition of \(\left\\{x_{1}, \ldots, x_{10}\right\\}\) into eight subsets. For example, the partition \(\left\\{x_{1}\right\\},\left\\{x_{2}\right\\},\left\\{x_{3}, x_{4}\right\\}\left\\{x_{5}\right\\},\left\\{x_{6}\right\\},\) \(\left\\{x_{7}, x_{8}\right\\}\left\\{x_{9}\right\\},\left\\{x_{10}\right\\}\) would be represented as $$x_{1}\left|x_{2}\right| x_{3} x_{4}\left|x_{5}\right| x_{6}\left|x_{7} x_{8}\right| x_{9} \mid x_{10}$$ Thus the solution to the problem is \(C(9,7)\).

Refer to \(a\) bag containing 20 balls-six red, six green, and eight purple. In how many ways can we select five balls if balls of the same color are considered identical?
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