Counting Relations

A relation on A is any subset of the Cartesian product A×A, which has 3×3 = 9 ordered pairs, so without constraints there are 2^9 relations. Reflexivity forces the three diagonal pairs (a,a),(b,b),(c,c) to be present in every relation, removing all freedom for those pairs. The remaining 9 - 3 = 6 off-diagonal pairs may each be included or excluded freely, giving 2^6 = 64 reflexive relations. This is a standard JEE Advanced counting pattern: fix mandatory pairs and count free choices on the rest. Option 2^9 counts all relations, ignoring the reflexive constraint that forces the diagonal pairs. Option 2^8 wrongly frees one of the three forced diagonal pairs. Option 2^3 counts only the choices on the diagonal, the exact opposite of what reflexivity allows. Thus the count is 2^6 = 64. Plausibility check: since reflexivity fixes 3 of 9 pairs, the exponent must drop from 9 to 6, and 2^6 indeed equals 64, consistent with the free off-diagonal choices. The same fix-the-forced-pairs-and-count-the-rest logic extends to counting symmetric or antisymmetric relations, where off-diagonal cells must be paired or restricted appropriately before counting free choices. Mastering precisely which ordered pairs a property forces, and which remain independent binary decisions, is the recurring competency that these relation-counting problems are designed to assess across many variations.