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📅 May 4, 2026 👤 scraface

May 04, 2026

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Choose two words from remaining 3 to have frequency 1: $\binom{3}{2} =...
The last word has frequency 0
For each such assignment, number of sequences: $ \frac{4!}{2!1!1!} = 1...
So total favorable: $4 \times 3 \times 12 = 144$ — same as before.
So yes, 144 favorable.
Total: $4^4 = 256$
So probability: $ \frac{144}{256} = \frac{9}{16} $? Wait: $144 \div 16...
But $ \frac{144}{256} = \frac{9}{16} $? $144 \div 16 = 9$, $256 \div 1...
But $ \frac{9}{16} $ is greater than 0.5, but is it reasonable? Let’s...
Alternatively, use known results: the number of sequences of length 4...

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