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📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 Set $ q_i = p_i - (i-1) $, then $ q_1 < q_2 < q_3 $ in $ \{1,\dots,3\} $, since $ p_3 \le 8 $, $ q_3 \le 8 - 2 = 6 $, but $ q_1 \ge 1 $, $ q_3 \ge q_2 + 2 \ge 1+2=3 $, $ q_3 \le 3 + 2 = 5 $? Let's see:

📌 $ q_3 = p_3 - 2 \le 8 - 2 = 6 $, but with $ q_1 \ge 1 $, $ q_2 \ge q_1 + 1 $, $ q_3 \ge q_2 + 1 $, and $ q_3 \ge q_1 + 2 $. With 3 variables, minimum span is 3 (e.g., 1,3,5).

📌 Maximum max position is 8, but with 3 non-adjacent in 5 positions, the last can be at most 5 if first at 1,3,5 — or 1,3,6 or 1,3,7,8—but 8 ≥ 6, so 1,3,6 or 1,3,7, or 1,3,8, or 1,4,6, etc.

📌 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.

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