`S:{1,2,3.................,10}`, M is the set of all subsets of S.
Let `R:MrightarrowM` be a relation such that `(A,B)inR"RightarrowAcapB=phi` then relation R is:

Options

A. Symmetric
B. Transitive
C. Reflexive
D. None

Answer

To solve the problem, we need to analyze the relation \( R \) defined on the set of all subsets \( M \) of the set \( S = \{1, 2, 3, \ldots, 10\} \). The relation \( R \) is defined such that \( (A, B) \in R \) if and only if \( A \cap B = \emptyset \), meaning that sets \( A \) and \( B \) are disjoint.

### Step-by-Step Solution:

1. **Understanding the Relation**:
- The relation \( R \) is defined on the power set \( M \) of \( S \).
- For any two subsets \( A \) and \( B \) in \( M \), \( (A, B) \in R \) if they have no elements in common, i.e., \( A \cap B = \emptyset \).

2. **Checking Reflexivity**:
- A relation is reflexive if every element is related to itself. That is, for all \( A \in M \), \( (A, A) \in R \) should hold.
- Here, \( A \cap A = A \), which is not equal to \( \emptyset \) unless \( A \) is the empty set.
- Therefore, \( R \) is **not reflexive**.

3. **Checking Symmetry**:
- A relation is symmetric if whenever \( (A, B) \in R \), then \( (B, A) \in R \) also holds.
- If \( A \cap B = \emptyset \), then it follows that \( B \cap A = \emptyset \) as well.
- Thus, \( R \) is **symmetric**.

4. **Checking Transitivity**:
- A relation is transitive if whenever \( (A, B) \in R \) and \( (B, C) \in R \), then \( (A, C) \in R \) must also hold.
- Suppose \( A \cap B = \emptyset \) and \( B \cap C = \emptyset \). However, \( A \cap C \) could potentially have elements in common.
- For example, let \( A = \{1\}, B = \{2\}, C = \{1\} \). Here, \( A \cap B = \emptyset \) and \( B \cap C = \emptyset \), but \( A \cap C \neq \emptyset \).
- Therefore, \( R \) is **not transitive**.

5. **Conclusion**:
- The relation \( R \) is symmetric but neither reflexive nor transitive.
- Hence, the correct answer is that the relation \( R \) is **symmetric**.