Problem 41 - Algebra for College Students

Question

How many four-element subsets containing A or B but not both \(A\) and \(B\) can be formed from the set \(\\{A, B, C\), D, E, F, G\\}? 20

Step-by-Step Solution

Answer

20 subsets.

Step 1: Understand the Problem

We need to form subsets of size 4 from the given set of 7 elements, ensuring each subset contains either the element A or B, but not both.

Step 2: Calculate Subsets Containing A but Not B

Consider subsets that contain A but do not contain B. We have the remaining elements C, D, E, F, and G to choose from. We need 3 more elements to complete our subset, so we calculate the number of ways to choose 3 elements from these 5:\[\binom{5}{3} = 10\]

Step 3: Calculate Subsets Containing B but Not A

Now consider subsets that contain B but do not contain A. Again, we have the same 5 elements C, D, E, F, and G to choose from. We need 3 more elements to complete our subset, so the calculation is similar:\[\binom{5}{3} = 10\]

Step 4: Add the Results

Add the number of subsets that contain A but not B to those that contain B but not A:\[10 + 10 = 20\]

Step 5: Conclusion

The total number of four-element subsets containing either A or B, but not both, is 20.

Key Concepts

Subsets

In combinatorics, a subset is any selection of elements from a given set. Subsets can include any number of elements, from no elements at all (the empty set) to the entire set. For example, if you have a set \(\{1, 2, 3\}\), its subsets include \(\{\}\), \(\{1\}\), \(\{2\}\), \(\{3\}\), \(\{1, 2\}\), \(\{1, 3\}\), \(\{2, 3\}\), and \(\{1, 2, 3\}\).
In this exercise, we focus on four-element subsets of a seven-element set \(\{A, B, C, D, E, F, G\}\). The key requirement is that each subset contains either \(A\) or \(B\) but not both. This problem illustrates how conditions can limit the number of valid subsets. An effective strategy is to divide the problem into smaller parts, calculate for each specific condition, and then combine the results to find the total possible subsets.

Permutations

Permutations involve the arrangement of elements in a specific order. Unlike with subsets, order is important in permutations. For instance, if you have a set \(\{A, B, C\}\), arranging two elements could result in \(AB\), \(AC\), \(BA\), \(BC\), \(CB\), or \(CA\).
In this exercise, however, we focus on subsets, where the order of elements in a subset does not matter. Therefore, we do not need to calculate permutations here. However, understanding permutations is crucial in other combinatorial problems where the order in which elements appear is significant, such as arranging books on a shelf or creating a sequence from a set of notes.

Binomial Coefficients

Binomial coefficients allow us to determine the number of ways to choose a specific number of elements from a larger set, regardless of the order in which they are chosen. This is represented as \(\binom{n}{k}\), where \(n\) is the total number of elements, and \(k\) is the number we want to choose.
In our example, we use the binomial coefficient \(\binom{5}{3}\) for choosing three additional elements from five options \(\{C, D, E, F, G\}\) to create a four-element subset, including \(A\) or \(B\). The result is 10, which signifies 10 different ways to complete each three-element combination with either \(A\) or \(B\). The final answer combines both scenarios (containing \(A\) but not \(B\), and vice versa) to give a total of 20 valid subsets.