The number of relations on the set $ A = \{1, 2, 3\} $ containing at most 6 elements including $ (1, 2) $, which are reflexive and transitive but not symmetric, is:
Show Hint
Correct Answer: 6
Solution and Explanation
We are given the set \( A = \{1, 2, 3\} \), and the relations \( (1, 1), (2, 2), (3, 3), (1, 2) \in R \). The remaining elements are: \[ (2, 1), (2, 3), (1, 3), (3, 1), (3, 2) \] We need to determine the number of relations on the set \( A \) containing at most 6 elements, which are reflexive and transitive but not symmetric.
Step 1: If the relation contains exactly 4 elements
The reflexive pairs \( (1, 1), (2, 2), (3, 3) \) must be included. We are required to select one additional element from the remaining ones: \[ (2, 1), (2, 3), (1, 3), (3, 1), (3, 2) \] Only 1 way is possible: We choose the pair \( (1, 2) \), which makes the relation transitive.
Thus, for 4 elements, there is exactly 1 way to form the relation.
Step 2: If the relation contains exactly 5 elements
Again, we must include \( (1, 1), (2, 2), (3, 3) \). Now, we must choose 2 additional elements from the remaining pairs: \[ (2, 1), (2, 3), (1, 3), (3, 1), (3, 2) \] The possible choices are: \( (1, 3) \) and \( (3, 2) \), or \( (3, 1) \) and \( (2, 3) \).
Thus, for 5 elements, there are 2 ways to form the relation.
Step 3: If the relation contains exactly 6 elements
In this case, we must include all reflexive pairs \( (1, 1), (2, 2), (3, 3) \) and all the other pairs: \[ (2, 1), (2, 3), (1, 3), (3, 1), (3, 2) \] There are 3 possible ways to form a 6-element relation: 1. \( \{(2, 3), (1, 3), (3, 2), (1, 1), (2, 2), (3, 3)\} \) 2. \( \{(2, 3), (3, 1), (1, 3), (3, 2), (1, 1), (2, 2)\} \) 3. \( \{(3, 2), (1, 3), (3, 1), (2, 1), (1, 1), (2, 2)\} \)
Thus, for 6 elements, there are 3 ways to form the relation.
Final Answer: Total number of ways = \( 1 + 2 + 3 = 6 \).
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