First, arrange the letters $A, A, I, N$. There are $\binom{4}{2}$ ways to choose the positions of the $A$s and $2!$ ways to arrange the remaining letters. Doing so creates five spaces in which the $S$s may be placed, three between successive letters and two at the ends of the row. For instance, if the arrangement is $AAIN$, we have $$\square A \square A \square I \square N \square$$ We have five choices where to place the double $S$. Once we choose its position, there are four empty spaces left in which to place the remaining $S$s. In order to separate them, we must choose two of these four spaces, which can be done in $\binom{4}{2}$ ways. Hence, the number of admissible arrangements is $$\binom{4}{2}2!\binom{5}{1}\binom{4}{2} = 360$$