Classic olympiad problem · Combinatorics
Prove that binomn0 + binomn1 + … + binomnn = 2^n by a counting…
Prove that by a counting argument, not by the binomial theorem.
This is a proof problem — the answer is an argument, not a number. The solution is deliberately not posted here: reading a solution you didn't fight for teaches almost nothing. On Lemma you attempt it cold, take one of the 3 progressive hints only when genuinely stuck, then compare your proof against a full walkthrough and mark yourself with the same rubric a competition coordinator would use.
Train it on Lemma
This problem sits in the Olympiad Set — 81 real competition problems with progressive hints, full walkthroughs and marking rubrics. 78 lessons, 624 curated problems and unlimited generated practice at six difficulties. Free to start — no card, no trial clock.
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