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where does pato o'ward live - Redraw

📅 May 4, 2026 👤 scraface

May 04, 2026

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📌 Number of such sequences:

📌 Choose which word appears twice: 4 choices

📌 Choose which word appears once (different): 3 choices

📌 Choose which word appears once (different): 2 choices

📌 Choose which word is unused: 1 choice

📌 Number of arrangements: $ \frac{4!}{2!1!1!} = 12 $

📌 So total: $4 \times 3 \times 2 \times 12 = 288$ — wait, that’s larger than total outcomes — contradiction.

📌 Ah! Here’s the mistake: we don’t need to choose the second and third separately. Once we pick the word that appears twice (4 choices), and the other two distinct words from the remaining 3 ( $\binom{3}{2} = 3$), the frequencies are fixed: one appears twice, two appear once, and one is unused. But in the multinomial count, we are overcounting because once we fix the repeated word and the two single words, the configuration is fully determined.

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