JEE Advanced 2026 Maths: Counting Equivalence Relations | NCERT to JEE, Fastest Method

Count equivalence relations on {1, 2, …, 10} with exactly 42 ordered pairs — a JEE Advanced 2026 maths question, solved step by step from your NCERT Class 12 book, then the quickest way to do it in under three minutes. A teacher and Arjun, a Class 12 student, work through it together. Arjun asks the doubts every student has — why 5² and not ⁵C₂? when do I divide by 2!? — and makes the common mistakes out loud, so you never have to. What you will learn • An equivalence relation is the same as splitting the set into classes (NCERT Class 12, Chapter 1) • Pairs = sum of the squares of the class sizes — the grid picture • Counting splits: choose, multiply, divide only for equal classes • The steady method (biggest class first) and the fastest one (handshakes and triangle numbers) • The traps that give 1260, 1890 or 3780 instead of 2520 • A cousin from the same paper: functions f with g(f(x)) = 2x → one-one functions + inclusion–exclusion (answer 1860) • Counting reflexive and symmetric relations straight from the grid • Speed: two free checks (range and parity), target times, a timed drill Chapters 0:00 The question that looked like a gift 2:42 What NCERT already gives you 6:07 Idea 1: pairs are squares 9:15 Idea 2: counting the splits 12:16 Solving it, the steady way 16:05 The quickest way: handshakes 19:11 Cousins: one-one functions and more 23:33 Speed: three minutes, not ten 26:13 Ladder, exit check and recap Practise this and more — questions, worksheets and animated lessons for CBSE and JEE: https://qnvibe.in/app/ Subscribe:    / @qnvibe   #JEEAdvanced #JEEMaths #EquivalenceRelations #NCERT #CBSEClass12 #QnVibe