I am trying to do this question from a high school textbook. The answer given is 1058400, but I can't get this answer.
Here's what I've tried so far:
There's 11!/(2^3) arrangements without any restrictions
If the two As are adjacent, then there's 10!/(2^2) arrangements (you can swap the As, but since they are repeated there's no need to)
Then you can have AE, AI, EI adjacent. In each case this is (10!*2!)/(2^3), as the two vowels can be swapped around, but the As are repeated.
Subtracting these cases from the number of arrangements without any restrictions gives 1360800, which is not the answer.
Best Answer
trancelocation did exact calculations, but he confused to give the final answer.
Of course the number of all possible arrangements without any restrictions (taking into account that A,T,M appear twice) is: $\color{blue}{\frac{11!}{2! \cdot 2! \cdot 2!}}$. But it's irrelevant.
So the answer is:
(stating trancelocation's answer)
All together: $$\color{blue}{\frac{7!}{2! \cdot 2!} \cdot \binom{8}{4} \cdot \frac{4!}{2!} = 1,058,400}$$