How many vertices, edges, faces, etc.

Take any vertex of the hypercube, and ask how many k-V's it
participates in.  To make a k-V it needs to combine with k adjacent and
orthogonal vertices, and there are (nCk) distinct ways of doing this
[that is, choose k directions out of n possible ones].  Then multiply
by 2^n, the total number of vertices.  But this involves multiple
counting, since each k-V is shared by 2^k vertices.  So divide by 2^k,
and this yields the answer: (nCk)*2^{n-k}.


For example, 12d hypercube:

 0-v:   4,096
 1-v:  24,576
 2-v:  67,584
 3-v: 112,640
 4-v: 126,720
 5-v: 101,376
 6-v:  59,136
 7-v:  25,344
 8-v:   7,920
 9-v:   1,760
10-v:     264
11-v:      24
12-v:       1