Question:
Let the set of all relations $ R $ on the set $ \{a, b, c, d, e, f\} $, such that $ R $ is reflexive and symmetric, and $ R $ contains exactly 10 elements, be denoted by $ S $.
Then the number of elements in $ S $ is ___.
Let the set of all relations $ R $ on the set $ \{a, b, c, d, e, f\} $, such that $ R $ is reflexive and symmetric, and $ R $ contains exactly 10 elements, be denoted by $ S $.
Then the number of elements in $ S $ is ___.
Show Hint
In reflexive and symmetric relations, always fix the diagonal pairs first, then choose symmetric unordered off-diagonal pairs (each contributes 2 elements).
Updated On: May 19, 2025
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Correct Answer: 105
Solution and Explanation
Step 1: Understand the set and the relation
- The set is \( \{a, b, c, d, e, f\} \), which has 6 elements.
- A relation \( R \) on this set is a subset of the Cartesian product \( \{a, b, c, d, e, f\} \times \{a, b, c, d, e, f\} \), i.e., a set of ordered pairs \( (x, y) \) where \( x, y \in \{a, b, c, d, e, f\} \).
- The total number of possible ordered pairs is \( 6 \times 6 = 36 \).
- The relation \( R \) must be:
\qquad - Reflexive: For every element \( x \in \{a, b, c, d, e, f\} \), the pair \( (x, x) \) must be in \( R \). So, \( R \) must include \( (a, a), (b, b), (c, c), (d, d), (e, e), (f, f) \), which are 6 pairs.
\qquad - Symmetric: If \( (x, y) \in R \), then \( (y, x) \in R \). This means that off-diagonal pairs (where \( x \neq y \)) come in pairs: if \( (x, y) \) is in \( R \), so is \( (y, x) \).
- \( R \) contains exactly 10 elements (i.e., 10 ordered pairs).
- We need to find the number of elements in \( S \), where \( S \) is the set of all such relations \( R \). Step 2: Count the elements in \( R \)
- Since \( R \) is reflexive, it must contain the 6 diagonal pairs: \( (a, a), (b, b), (c, c), (d, d), (e, e), (f, f) \).
- This accounts for 6 of the 10 elements in \( R \).
- The remaining elements in \( R \) are off-diagonal pairs (i.e., pairs \( (x, y) \) where \( x \neq y \)).
- Since \( R \) has 10 elements total, the number of off-diagonal pairs is: \[ 10 - 6 = 4. \]
- Because \( R \) is symmetric, off-diagonal pairs come in pairs: if \( (x, y) \) is in \( R \), then \( (y, x) \) must also be in \( R \). So, 4 off-diagonal pairs correspond to choosing pairs \( \{x, y\} \) (where \( x \neq y \), and the pair \( \{x, y\} \) represents both \( (x, y) \) and \( (y, x) \)).
- Thus, the number of unordered pairs \( \{x, y\} \) (where \( x \neq y \)) is: \[ \frac{4}{2} = 2. \]
So, each relation \( R \) consists of:
- The 6 reflexive pairs (fixed due to reflexivity),
- 2 unordered pairs \( \{x, y\} \), each contributing the ordered pairs \( (x, y) \) and \( (y, x) \), for a total of 4 ordered pairs.
This gives a total of \( 6 + 4 = 10 \) ordered pairs, which satisfies the condition.
Step 3: Count the number of unordered pairs \( \{x, y\} \)
- We need to choose 2 unordered pairs \( \{x, y\} \) where \( x \neq y \), and \( x, y \in \{a, b, c, d, e, f\} \).
- The number of ways to choose an unordered pair \( \{x, y\} \) (where \( x \neq y \)) from a set of 6 elements is the number of ways to choose 2 elements from 6, which is given by the combination formula \( \binom{n}{k} \): \[ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15. \]
- So, there are 15 possible unordered pairs: \( \{a, b\}, \{a, c\}, \{a, d\}, \{a, e\}, \{a, f\}, \{b, c\}, \{b, d\}\)
\(, \{b, e\}, \{b, f\}, \{c, d\}, \{c, e\}, \{c, f\}, \{d, e\}, \{d, f\}, \{e, f\} \).
Step 4: Choose 2 unordered pairs for \( R \)
- We need to select 2 unordered pairs from these 15 possible pairs to form \( R \).
- The pairs must be distinct (e.g., we can’t choose \( \{a, b\} \) twice), because each pair \( \{x, y\} \) corresponds to the distinct ordered pairs \( (x, y) \) and \( (y, x) \), and \( R \) is a set (no repeated elements).
- The number of ways to choose 2 distinct unordered pairs from 15 is: \[ \binom{15}{2} = \frac{15 \times 14}{2 \times 1} = 105. \]
Step 5: Define \( S \) and find its size
- \( S \) is the set of all relations \( R \) on \( \{a, b, c, d, e, f\} \) that are reflexive, symmetric, and have exactly 10 elements.
- Each \( R \in S \) is uniquely determined by choosing 2 unordered pairs \( \{x, y\} \), then including the corresponding ordered pairs \( (x, y) \) and \( (y, x) \), along with the 6 reflexive pairs.
- From Step 4, the number of ways to choose 2 unordered pairs is 105.
- Thus, the number of such relations \( R \)—which is the number of elements in \( S \)—is 105.
Final Answer: The number of elements in \( S \) is \( \boxed{105} \). Correct Answer Correct Answer: 105
- The set is \( \{a, b, c, d, e, f\} \), which has 6 elements.
- A relation \( R \) on this set is a subset of the Cartesian product \( \{a, b, c, d, e, f\} \times \{a, b, c, d, e, f\} \), i.e., a set of ordered pairs \( (x, y) \) where \( x, y \in \{a, b, c, d, e, f\} \).
- The total number of possible ordered pairs is \( 6 \times 6 = 36 \).
- The relation \( R \) must be:
\qquad - Reflexive: For every element \( x \in \{a, b, c, d, e, f\} \), the pair \( (x, x) \) must be in \( R \). So, \( R \) must include \( (a, a), (b, b), (c, c), (d, d), (e, e), (f, f) \), which are 6 pairs.
\qquad - Symmetric: If \( (x, y) \in R \), then \( (y, x) \in R \). This means that off-diagonal pairs (where \( x \neq y \)) come in pairs: if \( (x, y) \) is in \( R \), so is \( (y, x) \).
- \( R \) contains exactly 10 elements (i.e., 10 ordered pairs).
- We need to find the number of elements in \( S \), where \( S \) is the set of all such relations \( R \). Step 2: Count the elements in \( R \)
- Since \( R \) is reflexive, it must contain the 6 diagonal pairs: \( (a, a), (b, b), (c, c), (d, d), (e, e), (f, f) \).
- This accounts for 6 of the 10 elements in \( R \).
- The remaining elements in \( R \) are off-diagonal pairs (i.e., pairs \( (x, y) \) where \( x \neq y \)).
- Since \( R \) has 10 elements total, the number of off-diagonal pairs is: \[ 10 - 6 = 4. \]
- Because \( R \) is symmetric, off-diagonal pairs come in pairs: if \( (x, y) \) is in \( R \), then \( (y, x) \) must also be in \( R \). So, 4 off-diagonal pairs correspond to choosing pairs \( \{x, y\} \) (where \( x \neq y \), and the pair \( \{x, y\} \) represents both \( (x, y) \) and \( (y, x) \)).
- Thus, the number of unordered pairs \( \{x, y\} \) (where \( x \neq y \)) is: \[ \frac{4}{2} = 2. \]
So, each relation \( R \) consists of:
- The 6 reflexive pairs (fixed due to reflexivity),
- 2 unordered pairs \( \{x, y\} \), each contributing the ordered pairs \( (x, y) \) and \( (y, x) \), for a total of 4 ordered pairs.
This gives a total of \( 6 + 4 = 10 \) ordered pairs, which satisfies the condition.
Step 3: Count the number of unordered pairs \( \{x, y\} \)
- We need to choose 2 unordered pairs \( \{x, y\} \) where \( x \neq y \), and \( x, y \in \{a, b, c, d, e, f\} \).
- The number of ways to choose an unordered pair \( \{x, y\} \) (where \( x \neq y \)) from a set of 6 elements is the number of ways to choose 2 elements from 6, which is given by the combination formula \( \binom{n}{k} \): \[ \binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15. \]
- So, there are 15 possible unordered pairs: \( \{a, b\}, \{a, c\}, \{a, d\}, \{a, e\}, \{a, f\}, \{b, c\}, \{b, d\}\)
\(, \{b, e\}, \{b, f\}, \{c, d\}, \{c, e\}, \{c, f\}, \{d, e\}, \{d, f\}, \{e, f\} \).
Step 4: Choose 2 unordered pairs for \( R \)
- We need to select 2 unordered pairs from these 15 possible pairs to form \( R \).
- The pairs must be distinct (e.g., we can’t choose \( \{a, b\} \) twice), because each pair \( \{x, y\} \) corresponds to the distinct ordered pairs \( (x, y) \) and \( (y, x) \), and \( R \) is a set (no repeated elements).
- The number of ways to choose 2 distinct unordered pairs from 15 is: \[ \binom{15}{2} = \frac{15 \times 14}{2 \times 1} = 105. \]
Step 5: Define \( S \) and find its size
- \( S \) is the set of all relations \( R \) on \( \{a, b, c, d, e, f\} \) that are reflexive, symmetric, and have exactly 10 elements.
- Each \( R \in S \) is uniquely determined by choosing 2 unordered pairs \( \{x, y\} \), then including the corresponding ordered pairs \( (x, y) \) and \( (y, x) \), along with the 6 reflexive pairs.
- From Step 4, the number of ways to choose 2 unordered pairs is 105.
- Thus, the number of such relations \( R \)—which is the number of elements in \( S \)—is 105.
Final Answer: The number of elements in \( S \) is \( \boxed{105} \). Correct Answer Correct Answer: 105
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