Diagonals Formula Derivation: nC2 – n Simplified to n(n-3)/2 | Polygon Trick

Ever wondered how the polygon diagonals formula n(n-3)/2 actually comes from nC2 – n? In this clip, we don't just state the formula — we derive it step by step, right from first principles. Starting with nC2 (the combination formula), we break down n! as n × (n-1) × (n-2)!, cancel out the (n-2)! terms, and simplify down to n(n-1)/2. From there, we subtract n (since nC2 counts the sides of the polygon too, which aren't diagonals), take the LCM to combine terms into a single fraction, and simplify n² - n - 2n into n² - 3n. Finally, by taking n common, we arrive at the clean, elegant formula: n(n-3)/2. To make sure the formula actually works, we verify it manually for a triangle, quadrilateral, pentagon, and hexagon — cross-checking the algebraic answer against the actual count of diagonals drawn on each polygon. For the hexagon specifically, we carefully count all 9 diagonals on the diagram to confirm the formula holds true, catching a commonly missed diagonal in the process. This is a must-watch if you've ever memorized the diagonals formula without understanding where it comes from, or if you keep confusing nC2 with the actual diagonal count. Perfect for students revising Permutations & Combinations, preparing for board exams, or brushing up on mensuration and geometry-based combinatorics questions. By the end of this clip, you'll be able to derive n(n-3)/2 from scratch, explain why we subtract n from nC2, and verify the diagonal count for any polygon confidently — without rote-learning the formula. Watch till the end for the complete step-by-step algebraic simplification and the hexagon diagonal verification!