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### Problem Summary
We are given:
1. A set \( X = \{a, b, c, d, e, f\} \), which has 6 elements.
2. A symmetric relation \( R \) on \( X \), meaning that if \( (x, y) \in R \), then \( (y, x) \in R \) as well.
3. We need to find:
   - The number of symmetric relations on \( X \).
   - The number of those symmetric relations that are also reflexive.

### Solution Steps

1. **Total Number of Pairs**: Since \( X \) has 6 elements, there are \( 6 \times 6 = 36 \) possible ordered pairs \( (x, y) \) for \( x, y \in X \).

2. **Symmetric Condition**: To create a symmetric relation, if \( (x, y) \in R \), then \( (y, x) \) must also be in \( R \). Therefore:
   - **Diagonal Pairs**: There are 6 diagonal pairs \((a, a), (b, b), (c, c), (d, d), (e, e), (f, f)\), which are naturally symmetric (since they are of the form \((x, x)\)). We can choose whether or not to include each of these pairs independently.
   - **Off-Diagonal Pairs**: For each pair \((x, y)\) where \( x \neq y \), we can either include both \((x, y)\) and \((y, x)\) in the relation or exclude both. There are \( \binom{6}{2} = 15 \) such pairs (since there are 15 unique pairs of distinct elements in a set of 6).

3. **Counting Symmetric Relations**:
   - Each of the 6 diagonal pairs can be independently included or excluded, giving \( 2^6 = 64 \) choices.
   - Each of the 15 off-diagonal pairs also has 2 choices: include both \((x, y)\) and \((y, x)\) or exclude both. This gives \( 2^{15} \) choices for the off-diagonal pairs.
   - Thus, the total number of symmetric relations on \( X \) is \( 2^6 \times 2^{15} = 2^{21} = 2,097,152 \).

4. **Counting Reflexive and Symmetric Relations**:
   - For a relation to be reflexive, all diagonal pairs \((a, a), (b, b), (c, c), (d, d), (e, e), (f, f)\) must be included.
   - This means we have no choice for the 6 diagonal pairs—they must all be included.
   - We still have \( 2^{15} \) choices for the off-diagonal pairs (as we did for symmetric relations).
   - Therefore, the number of symmetric and reflexive relations is \( 2^{15} = 32,768 \).

### Final Answer
1. **Total symmetric relations**: \( 2^{21} = 2,097,152 \).
2. **Symmetric and reflexive relations**: \( 2^{15} = 32,768 \).

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### Related Questions

1. What is the difference between reflexive, symmetric, and transitive properties in relations?
2. How many symmetric relations are there on a set with \( n \) elements?
3. What changes if we also require the relation to be transitive?
4. How would the answer change if the set \( X \) had 5 elements instead of 6?
5. Can a relation be symmetric but not reflexive? Provide examples.

#### Tip:
For any set with \( n \) elements, a symmetric relation can be efficiently counted by separately considering diagonal and off-diagonal elements.

We say that a relation R on a set X is symmetric if (x, y) ∈ R implies (y, x) ∈ R for all x, y ∈ X. If X = {a, b, c, d, e, f}, how many symmetric relations are there on X? How many of these are reflexive?

Explore how to calculate the number of symmetric relations on a set X with 6 elements, as well as the count of those that are also reflexive. This solution involves combinatorial analysis and the properties of symmetric and reflexive relations in set theory.

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