So total: $4 \times 3 \times 2 \times 12 = 288$ — wait, that’s larger than total outcomes — contradiction.
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- Choose which word appears once (different): 2 choices
- Choose which word is unused: 1 choice
- Number of arrangements: $ \frac{4!}{2!1!1!} = 12 $
- 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.
- So better: number of ways to assign frequencies: one category with 2, two categories with 1, one with 0.
- Number of ways to assign the roles:
