Question

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:

Hint:

When dealing with reflexive and transitive relations, be sure to include the required reflexive pairs and check for transitivity by including necessary pairs. Avoid including pairs that would make the relation symmetric if the condition specifies it should not be symmetric.

Answer 1

Correct Answer: 6

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 = \( 1 + 2 + 3 = 6 \).

Answer 2

Given set: \[ A = \{1, 2, 3\} \] Relation \( R \) contains: \[ (1,1), (2,2), (3,3), (1,2) \in R \] Remaining possible ordered pairs are: \[ (2,1), (2,3), (1,3), (3,1), (3,2) \] --- (1) If the relation contains **exactly 4 elements**, there is only **1 way**. --- (2) If the relation contains **exactly 5 elements**, possible extra pairs can be: \[ (1,3), (3,2) \] Hence, **2 ways**. --- (3) If the relation contains **exactly 6 elements**, it can be any of the following: \[ (2,3), (1,3); \quad (1,3), (3,2); \quad (3,1), (3,2) \] Thus, **3 ways**. --- Therefore, \[ \text{Total number of ways} = 1 + 2 + 3 = 6 \] \[ \boxed{\text{Total = 6 ways}} \]