Solution:

Given such a $k$-configuration, we can write out all the elements of one of the $k$-element subsets, then all the elements of another subset, and so forth, eventually obtaining an ordering of all $n$ elements of $A$. Conversely, given any ordering of the elements of $A$, we can construct a consistent $k$-configuration of order 1 from it by grouping together the first $k$ elements, then the next $k$ elements, and so forth. In fact, each consistent $k$-configuration of order 1 corresponds to $(n / k)! (k!)^{n / k}$ different such orderings, since the elements of $A$ within each of the $n / k$ $k$-element subsets can be ordered in $k!$ ways, and the various subsets can also be ordered with respect to each other in $(n / k)!$ different ways. Thus, since there are $n!$ orderings of the elements of $A$, we get
$$\frac{n!}{(n / k)! (k!)^{n / k}}$$
different consistent $k$-configurations of order 1.