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📅 April 30, 2026 👤 scraface

Apr 30, 2026

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📌 \frac{4!}{2!1!1!0!} = 12

📌 For each such frequency pattern, number of ways to assign which word has which frequency.

📌 Choose the word with frequency 2: 4 choices

📌 Choose two words from remaining 3 to have frequency 1: $\binom{3}{2} = 3$

📌 The last word has frequency 0

📌 For each such assignment, number of sequences: $ \frac{4!}{2!1!1!} = 12 $

📌 So total favorable: $4 \times 3 \times 12 = 144$ — same as before.

📌 So yes, 144 favorable.

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