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First separating the word estate into vowels and non vowels gives $EAE$ for the vowels and $S,T,T$ for the non-vowels. I interpreted this as a group of 4 where there's two T's resulting in $\frac{4!}{2!}$ but since the vowels can also be permuted and there's two E's this results in #permutations = $\frac{4!}{2!}\cdot\frac{3!}{2!}=3!3!=36$.

The correct answer is $180$ so I'm off by a factor of 5 although I have no clue where it comes from. If someone could point out the flaw in my reasoning that would be great.

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  • $\begingroup$ I suppose my answer key is incorrect then, unfortunate. $\endgroup$ – Craig Dec 13 '18 at 8:09
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Your result is correct. If all the vowels are adjacent we have three possible vowel-blocks: $AEE$, $EAE$ and $EEA$. Then the total number of arrangents of $S$, $T$, $T$ and the vowel-block ($4$ elements with a double letter) is $$3\cdot \frac{4!}{2!}=36$$

Note that the total number of anagrams of the word $ESTATE$ with no restrictions is $$\frac{6!}{2!2!}=180.$$

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