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📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📝 Recent Posts

We need to place 4 H’s and 3 P’s in 7 positions, such that no two H’s...
First, place the 3 P’s. This creates 4 possible gaps (before, between,...
\_ P \_ P \_ P \_ → 4 gaps
We need to place 4 H’s into these 4 gaps, with **at most one H per gap...
There is only **one** way to choose positions: H in each gap → so the...
H P H P H P H → i.e., H P H P H P H
But wait — this places H in all 4 gaps, but total positions:
Each gap holding one H gives:
positions: H (1), P (2), H (3), P (4), H (5), P (6), H (7) → exactly 7...
So the structure is uniquely determined: H in positions 1,3,5,7 and P...

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