Let A={1,2,3}. Then number of relations containing (1,2) and (1,3) which are reflexive and symmetric but not transitive is
1
2
3
4
The given set is A = {1, 2, 3}.
The smallest relation containing (1, 2) and (1, 3) which is reflexive and symmetric,
but not transitive is given by: R = {(1, 1), (2, 2), (3, 3), (1, 2), (1, 3), (2, 1), (3, 1)}
This is because relation R is reflexive as (1, 1), (2, 2), (3, 3) ∈ R.
Relation R is symmetric since (1, 2), (2, 1) ∈R and (1, 3), (3, 1) ∈R.
But relation R is not transitive as (3, 1), (1, 2) ∈ R, but (3, 2) ∉ R.
Now, if we add any two pairs (3, 2) and (2, 3) (or both) to relation R, then relation R will
become transitive.
Hence, the total number of desired relations is one.
The correct answer is A (1)
A school is organizing a debate competition with participants as speakers and judges. $ S = \{S_1, S_2, S_3, S_4\} $ where $ S = \{S_1, S_2, S_3, S_4\} $ represents the set of speakers. The judges are represented by the set: $ J = \{J_1, J_2, J_3\} $ where $ J = \{J_1, J_2, J_3\} $ represents the set of judges. Each speaker can be assigned only one judge. Let $ R $ be a relation from set $ S $ to $ J $ defined as: $ R = \{(x, y) : \text{speaker } x \text{ is judged by judge } y, x \in S, y \in J\} $.
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: