JEE Relations Trick: Reflexive + Transitive but NOT Symmetric 🔥
❓ Question The number of relations on the set A = { 1 , 2 , 3 } , A=\{1,2,3\}, containing at most 6 elements , including ( 1 , 2 ) (1,2) , which are Reflexive Transitive But NOT symmetric , is equal to ? 🖼️ Question Image ✍️ Short Solution This is a logic-heavy JEE question . No brute-force counting — property-by-property filtering is the key. We proceed step by step. 🔹 Step 1 — Total possible ordered pairs For set A = { 1 , 2 , 3 } A=\{1,2,3\} : A × A = { ( 1 , 1 ) , ( 1 , 2 ) , ( 1 , 3 ) , ( 2 , 1 ) , ( 2 , 2 ) , ( 2 , 3 ) , ( 3 , 1 ) , ( 3 , 2 ) , ( 3 , 3 ) } A\times A=\{(1,1),(1,2),(1,3),(2,1),(2,2),(2,3),(3,1),(3,2),(3,3)\} Total = 9 ordered pairs 🔹 Step 2 — Apply the reflexive condition A relation is reflexive iff: ( 1 , 1 ) , ( 2 , 2 ) , ( 3 , 3 ) ∈ R (1,1),(2,2),(3,3)\in R So these 3 elements are compulsory . 🔹 Step 3 — Mandatory given element The question says: ( 1 , 2 ) ∈ R (1,2)\in R So compulsory elements so far: { ( 1 , 1 ) , ( 2 ,...