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Fotos auf art2hotel.com. - SUpost

📅 July 6, 2026 👤 scraface

Jul 06, 2026

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📌 But now, the remaining 5 positions must hold 2 A’s and 3 G’s, with no two A’s adjacent and no two G’s adjacent.

📌 But as noted, placing 3 G’s in 5 positions with no two adjacent is only possible if they are in positions with gaps.

📌 Let’s compute how many ways to place 3 non-adjacent G’s in 5 positions.

📌 Let the positions be $ p_1 < p_2 < p_3 $, $ p_{i+1} \ge p_i + 2 $.

📌 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.

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