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Home » Breaking: Hoyoverse Log In Trap That’s Hiding Golusers Every Day! - Abbey Badges

Breaking: Hoyoverse Log In Trap That’s Hiding Golusers Every Day! - Abbey Badges

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Actually, in a subset of 5 consecutive positions, number of ways to choose 3 non-adjacent is equivalent to $ \binom{5 - 3 + 1}{3} = \binom{3}{3} = 1 $? That formula is for linear arrangement with no adjacency.

📌 Standard stars and bars: number of ways to choose k non-consecutive positions from n is $ \binom{n - k + 1}{k} $.

📌 Here, n = 5, k = 3, so $ \binom{5 - 3 + 1}{3} = \binom{3}{3} = 1 $.

📌 So only **one** way to place 3 non-adjacent G’s in 5 specific positions — only if they are in positions that can be mapped to {1,2,3,4,5} with gaps.

📌 But it’s only possible if the 5 positions allow a 3-element non-adjacent subset — which requires that the positions span at least 5 positions (e.g., 1,3,5), and in any 5 consecutive positions, there is exactly **one** such set? Not quite — for positions {1,2,3,4,5}, non-adjacent triples: 1,3,5 only? But 1,4,5 has adjacent, 2,4,5 has adjacent. Yes, only {1,3,5} if starting at 1, but depending on layout.

📌 Actually, the number is not always 1. For example, {1,3,4} has adjacent 3-4, invalid. So only triples with min gap 2.

📌 In a sequence of 5 positions, the number of 3-element subsets with no two adjacent is $ \binom{3}{3} = 1 $? No:

📌 General formula: $ \binom{n - k + 1}{k} = \binom{5 - 3 +

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