Problem 25 - Algebra for College Students

Question

How many three-element subsets containing the letter A can be formed from the set \(\\{A, B, C, D, E, F\\}\) ? 10

Step-by-Step Solution

Answer

10 subsets can be formed.

Step 1: Understanding the Problem

We need to find the number of three-element subsets from the given set \(\{A, B, C, D, E, F\}\) that contain the letter A. This means A must be in every subset, and we choose the remaining two elements from the other letters.

Step 2: Choosing Elements Besides A

After choosing A, we have 5 elements left: \(\{B, C, D, E, F\}\). We need to choose 2 more elements from this subset of 5 elements.

Step 3: Using Combinations to Find Subsets

The number of ways to choose 2 elements from 5 elements is given by the combination formula \(\binom{5}{2}\), which calculates the number of combinations of 5 elements taken 2 at a time.

Step 4: Calculating the Combinations

Calculate \(\binom{5}{2}\) using the formula: \[\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10\]

Step 5: Conclusion

There are 10 different three-element subsets containing the letter A that can be formed from the set \(\{A, B, C, D, E, F\}\).

Key Concepts

Subsets

In combinatorics, a subset is a collection of elements chosen from a larger set, where the order of the elements does not matter. When we talk about subsets, we are simply considering a group within a larger group. For instance, if you have a set \(\{A, B, C, D, E, F\}\), you can form smaller groups or subsets from this larger group. Each element of the original set can either be included in or excluded from a subset.

• A subset can contain no elements at all, which is known as the empty set.
• It can also consist of all elements from the original set, which makes it equal to the original set.

In this exercise, our task was specifically to find three-element subsets that include the letter A. This means that A is already included, and we select the remaining elements from the other letters in the set. Picking subsets efficiently becomes especially important when the set is large, hence the role of other counting techniques in combinatorics.

Combination Formula

The combination formula is crucial when it comes to counting subsets in combinatorics. The formula \(\binom{n}{r}\) helps you determine how many ways you can choose \(r\) elements from a set of \(n\) elements without considering the order of selection. The formula is expressed as:\[\binom{n}{r} = \frac{n!}{r!(n-r)!}\]

• \(n!\) stands for the factorial of \(n\). It is the product of all positive integers up to \(n\).
• Similarly, \(r!\) is the factorial of \(r\).

In our example, after including A as a fixed element in every subset, we were left to choose 2 additional elements from the remaining 5. The solution uses the combination formula to find out that there are \(\binom{5}{2}\) ways to choose these 2 elements, which equals 10 as calculated:\[\frac{5 \times 4}{2 \times 1} = 10\]This calculation shows us there are 10 different ways to select two additional letters to go with A.

Counting Principles

Counting principles are fundamental concepts in combinatorics that help us determine how many different ways an event can occur. These principles include basic rules and methods that simplify the counting process.Key principles:

• **Addition Principle**: If an event can occur in \(m\) ways and another mutually exclusive event can occur in \(n\) ways, then there are \(m + n\) ways for either event to happen.
• **Multiplication Principle**: If two events are independent and the first event can occur in \(m\) ways and the second can occur in \(n\) ways, then the events together can occur in \(m \times n\) ways.

In our subset problem, we're applying these principles specifically to determine the number of ways to choose elements. We multiply the number of ways to choose elements after fixing A, emphasizing independence in each choice. We calculate using the multiplication principle by selecting one element (A) and then the other two from what remains.