Symmetric relation

Step 1: In a symmetric relation, the (4) diagonal pairs are chosen independently. Step 2: The off-diagonal pairs form (\frac{4\cdot3}{2}=6) mirror-pair groups. Step 3: Total independent choices are (10), so the number of relations is (2^{10}).

Step 1: The (5) diagonal pairs are fixed, so they give no choice. Step 2: There are (\frac{5\cdot4}{2}=10) off-diagonal mirror groups. Step 3: Each group is either included fully or excluded, so the number is (2^{10}).

Step 1: A symmetric relation must contain ((b,a)) whenever ((a,b)) is present. Step 2: ((1,2)) and ((2,1)) are already paired, but ((2,3)) needs ((3,2)). Step 3: Adding only ((3,2)) makes the relation symmetric.

Step 1: If (a+b) is even, then (b+a) is also even because addition is commutative. Step 2: So ((a,b)\in R) implies ((b,a)\in R). Step 3: For such questions, reverse the pair and check whether the condition remains true.

Step 1: For ((2,1)), (2-1=1), so it belongs to the relation. Step 2: For the reverse pair ((1,2)), (1-2=-1), which is not positive. Step 3: One counterexample is enough to prove that a relation is not symmetric.

Step 1: ((a,b)\in R\cap S) means the pair is in both relations. Step 2: Since both are symmetric, ((b,a)) is also in both. Step 3: Hence ((b,a)\in R\cap S), so the intersection is definitely symmetric.

Step 1: In a symmetric relation, every pair appears with its reverse. Step 2: The inverse relation (R^{-1}) reverses all ordered pairs. Step 3: Reversing gives the same set of pairs, so (R^{-1}=R).

Step 1: Exactly (3) diagonal pairs can be chosen in (\binom{4}{3}=4) ways. Step 2: There are (6) independent off-diagonal mirror groups. Step 3: Therefore the total number is (4\cdot2^6).

Step 1: A relation is symmetric when its matrix is symmetric about the main diagonal. Step 2: Here (m_{13}=m_{31}=1), (m_{23}=m_{32}=1), and (m_{12}=m_{21}=0). Step 3: Hence the relation is symmetric.

Step 1: For a symmetric relation matrix, (m_{ij}=m_{ji}). Step 2: Here (m_{12}=1) but (m_{21}=0), while the other opposite entries match. Step 3: So changing only (m_{21}) to (1) makes the matrix symmetric.

Step 1: If (|a-b|=1), then (|b-a|=1) also. Step 2: Hence every pair has its reverse, so the relation is symmetric. Step 3: For ((a,a)), (|a-a|=0), so it is not reflexive.

Step 1: Numbers with the same remainder still have the same remainder after reversing their order. Step 2: Thus (a\equiv b \pmod{2}) implies (b\equiv a \pmod{2}). Step 3: Symmetry is exactly about checking this reverse pair.

Step 1: Since (1\mid2), ((1,2)\in R). Step 2: But (2\mid1) is false, so ((2,1)\notin R). Step 3: One such pair is enough to disprove symmetry.

Option A is correct. In a symmetric relation, each off-diagonal pair {a,b} must be selected as a complete two-way choice: either both (a,b) and (b,a) are included or neither is included. There are n(n−1)/2 such unordered pairs. Since diagonal pairs are forbidden and each off-diagonal pair can be chosen independently, the total number is 2^(n(n−1)/2).

Step 1: For (3) elements, there are (\frac{3\cdot4}{2}=6) independent groups. Step 2: If ((1,2)) is included, symmetry forces ((2,1)), fixing one group. Step 3: The remaining (5) groups are free, so the count is (2^5).

Step 1: There are (6) independent groups in total. Step 2: The group containing ((1,2)) is fixed as included, and the diagonal pair ((1,1)) is fixed as excluded. Step 3: So (4) groups remain free, giving (2^4).

For any relation, R⁻¹ is obtained by reversing ordered pairs. Since R is symmetric, reversing its pairs produces no new pairs and gives R⁻¹ = R. Substituting this into the expression yields R ∪ R⁻¹ = R ∪ R. The union of a set with itself is the set itself, so the expression equals R. It is not necessarily empty or universal.

Step 1: If ((a,b)\in R\cup R^{-1}), it belongs to (R) or (R^{-1}). Step 2: In either case, the reverse pair ((b,a)) belongs to (R\cup R^{-1}). Step 3: Therefore it is always symmetric.

The condition defining R is a+b=5. If (a,b) satisfies it, then reversing the components gives b+a=5 because addition is commutative; hence (b,a) also belongs to R, so R is symmetric. Reflexivity would require (a,a) for every a in A. But for a=1, 1+1≠5, and similarly the diagonal pairs are absent. Therefore R is symmetric but not reflexive.

Step 1: Changing the order in addition does not change the sum. Step 2: If (a+b) is odd, then (b+a) is also odd. Step 3: Hence every pair has its reverse, so (R) is symmetric.