To address this, we will evaluate the conditions for relation \( R \) on set \( A = \{0, 1, 2, 3, 4, 5\} \), where \( (x, y) \in R \) if and only if \(\max\{x, y\} \in \{3, 4\}\).
First, let's determine the pairs where \(\max\{x, y\} = 3\):
- If \( x = 3 \) or \( y = 3 \), then \(\max\{x, y\} = 3\). The applicable pairs are: \((3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (0, 3), (1, 3), (2, 3), (4, 3), (5, 3)\).
Next, identify the pairs where \(\max\{x, y\} = 4\):
- If \( x = 4 \) or \( y = 4 \), and \(\max\{x, y\} = 4\). The applicable pairs are: \((4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (0, 4), (1, 4), (2, 4), (3, 4), (5, 4)\).
Combining these, the complete set of ordered pairs in \( R \), without duplicates, is:
- \((3, 0), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (0, 3), (1, 3), (2, 3), (4, 0), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (0, 4), (1, 4), (2, 4), (5, 3), (3, 4)\)
The cardinality of \( R \) is \(20\), not 18. Consequently, Statement \( S_1 \) is false.
Next, we analyze whether relation \( R \) is symmetric, reflexive, or transitive:
- Symmetric: If \((x, y) \in R\), then \((y, x) \in R\). For every pair in \( R \), its inverse is also present. Therefore, the relation is symmetric.
- Reflexive: For all \( x \in A \), \((x, x) \in R\). This condition is not met, as pairs such as \( (0, 0), (1, 1), (2, 2) \) are absent from \( R \). Thus, the relation is not reflexive.
- Transitive: If \((x, y) \in R\) and \((y, z) \in R\), then \((x, z) \in R\). For instance, \((3, 0) \in R\) and \((0, 4) \in R\), but \((3, 4)\) is not in \( R \). Thus, the relation is not transitive.
Based on the preceding evaluation, Statement \( S_2 \), asserting that the relation is symmetric but neither reflexive nor transitive, is true.
Therefore, the correct conclusion is: only \( (S_2) \) is true.