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your letters are A, A, E, N, S, T, T, T. Find the probability that AAE is together in any order

We consider $AAE$ as one so $AAE \to O$

We have $O, N, S, T, T, T$, ways to arrange this is $6!/(3!) = 6*5*4 = 120$. Ways to arrange $O$ itself are $3$ so total our favored event is $120*3 = 360$.

Total possible: $8!/(2!3!) = 8*7*6*5*2 = 3360$

Thus $P = 360/3360$

Is this correct?

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    For the sake of computing probability, you can (and should IMO) count identical arrangements as unique. Thus, the probability is $\frac{3!\cdot(5+1)!}{(3+5)!}$. – 2017-01-20
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    This answer is the same as yours, but I'm not so sure that your method would work for any other combination of letters. – 2017-01-20
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    Yes you are correct (in this case). One follow-up question: what if you now have $3$ A's and $2$ O's? Then your method might not work immediately. In that case, how should one do it? – 2017-01-20
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    @k99731 I don't understand your question but if you have a question then post it as new question. – 2017-01-20
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    @miracle173 Forget about it then. Your approach is correct. – 2017-01-20

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Yes your answer is correct.

Taking the alphabets that comes together as a group is nice approach.

But if you write the ways of arranging O iteslf as $\frac{3!}{2!}$ It looks more understandable.