# Find the number of permutations of the word 'ESTATE' if all vowels must be adjacent

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.

• I suppose my answer key is incorrect then, unfortunate. – Craig Dec 13 '18 at 8:09

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 $$\frac{6!}{2!2!}=180.$$