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📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📝 Recent Posts

So after placing 3 non-adjacent C’s (20 ways), we have 5 remaining pos...
But wait: if we place 3 G’s in 5 positions with no two adjacent, is it...
Try: placing 3 G’s with no two adjacent in 5 positions.
Let the G’s occupy 3 positions such that no two are adjacent.
Then gaps: between G’s must be at least one non-G.
Minimum length for 3 non-adjacent G’s: G _ G _ G → 5 positions. Only o...
Positions: must satisfy $ g_2 \ge g_1 + 2 $, $ g_3 \ge g_2 + 2 $. The...
Only one such triplet of positions in 8? No — multiple.
But we already fixed C’s, and the 5 free positions are a subset of the...
So the 5 free positions may not even be contiguous.

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