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From remaining 3, choose 2 to occur once each: $\binom{3}{2} = 3$
The last one occurs once? No — we have only four positions. If one occurs twice, two others once, that’s 2+1+1 = 4, so the fourth is covered. So the fourth outcome (the unused one) doesn’t need to be chosen — it’s determined.
But when counting sequences, we need to assign positions.
Standard multinomial: number of sequences with frequencies (2,1,1,0) is:
\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 $
