You basically have it. For each string of beads, there are two possible permutations you can create (because you can start at either end of the string). Conversely, for every possible permutation, there is a unique string of beads that you can make which will produce that permutation. So if $N$ represents the number of possible strings of beads and $n!$ represents the number of possible permutations, you have shown that $2 N = n!$, so $N = \frac{n!}{2}$.
Edit: This is an attempt to make the argument clearer. You have a function which maps the set of all permutations of $1,2,\dots,n$ to the set of all strings of $n$ beads (the beads numbered $1,2,\dots,n$). Given a permutation $a_1 a_2 \cdots a_n$ (i.e. $a_i$ is the $i^{\text{th}}$ number in the permutation), then make a string of beads in the order $a_1, a_2, \dots, a_n$. This is a surjective function (for every string of beads, there is some permutation corresponding to it). In fact the preimage of any string of beads consists of two permutations since you either read the beads from left to right or right to left. Thus, you have a $2$-to-$1$ function from a set with $n!$ elements. The image set therefore has cardinality $n!/2$.