Question

A bookbinder is to bind 10 different books using red, green, or blue cloth for each book. How many ways can this be done if each color is to be used for: (a) At least one book (b) At least two books (c) With no restriction on the number of times a color may be used

Short answer

(a) 55,980 ways, (b) 40,620 ways, (c) 59,049 ways.

Step-by-step solution

Calculate total ways without restriction

Each of the 10 books can be bound in any of the three colors: red, green, or blue. Therefore, each book has 3 choices, leading to a total of \(3^{10}\) ways to bind all the books without any color restrictions.\[3^{10} = 59049\]

Apply Inclusion-Exclusion Principle for part (a)

For each color to be used at least once, we use the principle of inclusion-exclusion. First, use \(3^{10}\) total ways minus cases where at least one color is not used. Denote A, B, C as sets where red, green, blue are not used, respectively. Each set has \(2^{10}\) ways, and intersections of two sets (e.g., both red and green are not used) have \(1^{10}\) ways. Using inclusion-exclusion: \[\begin{align*} |A \cup B \cup C| &= 3 \times 2^{10} - 3 \times 1^{10} \ &= 3072 - 3 \ &= 3069 \end{align*}\] So, the number of ways with each color used at least once is \(3^{10} - |A \cup B \cup C| = 59049 - 3069 = 55980\).

Ensure at least two books for each color for part (b)

To ensure each color is used at least twice, subtract cases with one book in a particular color from results in Step 2. Calculate for each color being used only once (3 ways): choose one book for a fixed color (10 choices), and the remaining 9 are split among the other two colors: \(2^9\). This gives \[3 \times 10 \times 2^9 = 15360\] Subtract from the result of Step 2: \[55980 - 15360 = 40620\]

Check options for part (c) with no restrictions

With no restrictions on usage of colors, the total ways is simply \(3^{10}\) which was calculated in Step 1 as 59049.

Key concepts

Inclusion-Exclusion Principle

In combinatorics, the Inclusion-Exclusion Principle is a fundamental tool used to solve problems where the goal is to determine the number of elements in the union of several sets. This principle provides a systematic way to include and exclude overlapping sets in order to avoid double-counting.

Let's consider the step-by-step application of this principle in our exercise. The task is to find the number of ways to bind books using red, green, or blue cloth, so that each color is used at least once. Initially, we calculate the total number of ways by raising the number of choices (3 colors) to the power of the number of books (10), resulting in \(3^{10}\). This gives us the total ways without any restrictions.

To ensure each color is used at least once, we first exclude scenarios where a certain color isn't used. For instance, the sets denoted as \( A, B, \) and \( C \) represent cases where red, green, or blue are not used. Each of these has \(2^{10}\) ways since only the other two colors are available for each book. Furthermore, accounting for overlaps—like scenarios where exactly two colors aren't used, which happens in \(1^{10}\) way—allows the application of inclusion-exclusion:

• Sum the ways of individual exclusions (single colors not used).
• Subtract the overlaps (two colors not used simultaneously).

The final formula is \(3 \times 2^{10} - 3 \times 1^{10} = 3069\), which is then subtracted from the \(3^{10}\) total, ensuring no redundancy in colorless scenarios.

Permutations and Combinations

Permutations and combinations are central concepts in discrete mathematics that deal with counting and arranging objects. In our exercise, we're primarily concerned with combinations, as we are looking at different ways to "combine" colors and books without concern for order.

A permutation is an arrangement of elements where order matters. For example, arranging three books in different orders would be using permutations. However, in our problem, colors are a choice for each book, and the order of the binding doesn’t affect the outcome, which aligns more with combinations. Here, we are selecting from three colors (red, green, blue) for each book, and these selections form different combinations.

Combinations without any constraints are simpler, calculated by the number of ways to choose items without regard to order. Here, each book represents a choice of any of the three options independently. When calculating with no restrictions, this results in \(3^{10}\) choices possible, since each of the ten books can independently be any of the three colors. Contrast this with more restricted combinations where particular conditions must be met, such as each color being used at least once, requiring application of more complex formulas like the inclusion-exclusion principle.

Discrete Mathematics

Discrete mathematics is a branch of mathematics dealing with countable, distinct elements. It's incredibly useful in computer science and for solving real-life problems involving finite structures. Within this field, combinatorics is a key area that focuses on counting, arranging, and analyzing the patterns of elements.

Our exercise belongs squarely in this field, as it deals with a finite number of books and colors to arrive at solutions. Discrete mathematics provides the tools and theories, like permutations, combinations, and principles such as inclusion-exclusion, to systematically solve complex counting problems.

In discrete mathematics, each solution step is often a clear, logical progression, applying relevant theorems and principles. For instance:

• Starting with the calculation of \(3^{10}\) for total choices, we demonstrate basic operations on sets.
• The inclusion-exclusion principle is applied to manage overlapping sets, ensuring each possibility is appropriately accounted.

This precise, step-by-step methodology is a hallmark of discrete mathematics, enabling us to tackle and solve problems with clear, precise reasoning.