Let A be a set and let B be the set of all the subsets of A.
(a) Is it possible for B to contain exactly one set of cardinality 7?
If the answer is positive, what can you say about A? Justify your answers.
(b) Is it possible for B to contain exactly two sets of cardinality 7?
If the answer is positive, what can you say about A? Justify your answers

Let’s analyze the problem step by step.

### (a) Is it possible for \( B \) to contain exactly one set of cardinality 7?

#### **Step 1: Understanding the set \( B \)**
- \( A \) is a set, and \( B \) is the power set of \( A \). This means that \( B \) contains **all subsets of \( A \)**.
- The number of subsets of \( A \) is \( 2^{|A|} \), where \( |A| \) is the cardinality of \( A \) (i.e., the number of elements in \( A \)).
- Among these subsets, we are particularly interested in those subsets of cardinality 7.

#### **Step 2: Analyzing the problem**
For \( B \) to contain exactly one subset of cardinality 7, the original set \( A \) must have at least 7 elements. The number of subsets of a set \( A \) of cardinality \( n \), that have cardinality 7 is given by the binomial coefficient \( \binom{n}{7} \).

So we require:

\[
\binom{n}{7} = 1
\]

This condition implies that \( n \) must be exactly 7 because:

\[
\binom{n}{7} = 1 \quad \text{if and only if} \quad n = 7
\]

#### **Conclusion for (a)**
- **Yes**, it is possible for \( B \) to contain exactly one set of cardinality 7 if and only if \( A \) has exactly 7 elements. In this case, the only subset of cardinality 7 is \( A \) itself.
  
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### (b) Is it possible for \( B \) to contain exactly two sets of cardinality 7?

#### **Step 1: Reconsidering the number of subsets of cardinality 7**
For \( B \) to contain exactly two subsets of cardinality 7, we need:

\[
\binom{n}{7} = 2
\]

However, the binomial coefficient \( \binom{n}{7} \) only takes integer values and increases as \( n \) increases. The smallest values for \( \binom{n}{7} \) as \( n \) increases are as follows:

- \( \binom{7}{7} = 1 \)
- \( \binom{8}{7} = 8 \)
- \( \binom{9}{7} = 36 \)

There is no \( n \) such that \( \binom{n}{7} = 2 \). Therefore, it is **impossible** for \( B \) to contain exactly two subsets of cardinality 7.

#### **Conclusion for (b)**
- **No**, it is not possible for \( B \) to contain exactly two sets of cardinality 7.

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### Summary:

- **(a)** Yes, \( B \) can contain exactly one set of cardinality 7 if and only if \( A \) has exactly 7 elements.
- **(b)** No, it is not possible for \( B \) to contain exactly two sets of cardinality 7.

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Would you like more details on any part of the reasoning?

Here are some related questions for deeper understanding:
1. What is the power set of a set, and how is it calculated?
2. How does the binomial coefficient \( \binom{n}{k} \) relate to subsets of a given cardinality?
3. Can we calculate the exact number of subsets of a given size for any set?
4. How would the situation change if we asked about subsets of a different cardinality, say 5?
5. What is the total number of subsets in a set of cardinality \( n \)?

**Tip:** Understanding how binomial coefficients work is essential when analyzing problems involving subsets. These coefficients give us the number of ways to choose subsets of a specific size from a set.

Let A be a set and let B be the set of all the subsets of A.
(a) Is it possible for B to contain exactly one set of cardinality 7?
If the answer is positive, what can you say about A? Justify your answers.
(b) Is it possible for B to contain exactly two sets of cardinality 7?
If the answer is positive, what can you say about A? Justify your answers.

Explore whether a power set can contain exactly one or two subsets of cardinality 7. This problem involves set theory and combinatorics, using binomial coefficients to determine the conditions under which specific subsets exist. Suitable for undergraduate-level students.

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