AMC 12 Mock Exam 2 Problem 8

First arrange the four letters other than $\text{A}$, namely $\text{B},\text{N},\text{N},\text{S}$; there are $\frac{4!}{2!}=12$ arrangements. These four letters create $5$ gaps (including the two ends), and placing the three $\text{A}$'s in three different gaps is exactly what keeps them apart; there are $\binom53=10$ ways. The answer is $12\cdot10=\boxed{120}$.

(The total number of arrangements, with no restriction, is $\frac{7!}{3!\,2!}=420$.)