What is the number of $(n-r)$ dim faces for a $k$-ary $n$-dim cube ?
Definition of k-ary cube: In a $k$- ary $n$- cube , each node is identified by an $n$-bit base-$k$ address $b_{n − 1},...b_j,...b_0$ , and for each dimension $j =0,1,...n − 1$, the node is connected by edges to nodes with addresses $b_{n − 1},...,b_j ± 1 (mod k ),...b_0 $. For details check this paper's introduction
To put this thing simply: when we consider a unit square, or a unit cube or a unit n-dim cube each of it's vertices can be represented as: $\{0,1\}^n$. So this is an example of $k=2$. To generalize this idea we can assume $k$ as any +ve integer and is called as $k$-ary $n$-dim cube.
e.g: for $k=2, n = 3$ it's unit 3-d cube:
$(r=1)$the number of $3-1 = 2$ (dim)-faces = 6
$(r=2)$the number of $ 3-2 = 1$ (dim) faces i.e edges = 12
Please let me know if more clarification is needed.