\(\text{The way to look at this semi formally is to consider the different cases}\\ \text{There can be 0-2 pairs of letters chosen}\\ \text{with 0 pairs there are }\dbinom{7}{4}4! =840 \text{ unique 4 letter words}\\ \text{with 1 pair there are }3\dbinom{4}{2}\dbinom{6}{2}2!=540 \text{ unique 4 letter words}\\ \text{This last bit deserves some discussion}\\ \text{The 3 comes from the fact that 3 of the letters can be chosen as a pair, I, N, V}\\ \text{The }\dbinom{4}{2} \text{ comes from the fact that we select two slots from the 4 to place this pair}\\ \text{The }\dbinom{6}{2} \text{ comes from the fact that we can select 2 letters from 6 to complete the word}\\ \text{The 6 comes from removing the numbers that are the pair, and the }\\ \text{2nd copies of the two other numbers}\\ \text{Finally the }2! \text{ comes from the fact the these last two letters are distinct and their order matters}\)
\(\text{Finally with 2 pairs there are }\dbinom{3}{2}\dbinom{4}{2} =18 \\ \text{This is choosing two pairs from the 3 available and choosing 2 slots from the 4}\\ \text{adding all these up we get}\\ 840+540+18 = 1398 \)
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