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📅 May 18, 2026 👤 scraface

May 18, 2026

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📌 0000, 0001, 0010, 0100, 0101, 1000, 1010 → 7

📌 00000, 00001, 00010, 00100, 00101, 01000, 01001, 01010, 10000, 10001, 10010, 10100, 10101 → 13

📌 So \(a_5 = 13\)

📌 But this counts **classifications** — each position gets H or S, no two adjacent H’s.

📌 Since the substations are in a fixed line (distinct positions), and each is assigned one mode, the total number is exactly the number of such binary strings: 13.

📌 Wait — but earlier we computed \(\binom{n-k+1}{k} = \binom{5-4+1}{4} = \binom{2}{4} = 0\) — invalid.

📌 Mistake: formula \(\binom{n-k+1}{k}\) is for placing \(k\) non-adjacent indistinct items? No — standard formula for number of binary strings of length \(n\) with no two consecutive 1’s is \(F_{n+2}\), where \(F_1=1, F_2=1, F_3=2, F_4=3, F_5=5, F_6=8, F_7=13, F_8=21\)

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