What is the number of size 3 subsets of vertices of an n-dimensional hypercube such that the 3 vertices are pairwise a distance d apart? Equivalently, what is the number of size 3 subsets S of binary words of length n such that the Hamming distance between each pair of words in S is exactly k.
I think that if k is odd then there are no such subsets.  This is because if x,y,z are binary words then D(x,z) + e = D(x,y) + D(y,z) where e is an even nonnegative integer.  
If k = 2, for n = 1,2,...,8.  I have a brute force Mathematica code that returns 0,0,8,64,320,1280,4480,14336.
Is there a simple combinatorial argument to count these subsets of vertices?
Can someone verify the terms that I have given above for the case k=2?
 A: Here is an answer for $k=2$.  
For three binary words to have a Hamming distance of $k=2$, the letters of these three words have to coincide at $n-3$ positions. There are $2^{n-3}$ choices. 
For the remaining 3 positions, there are $n \choose 3$ many choices. 
Consider that one word has 3 binary letters (a,b,c) at these positions. Then, for pairwise distance of $k=2$,  the second word needs to have $(\bar a, \bar b, c)$ and the third word $(\bar a, b, \bar c)$ where the bars denote the opposite choice of the binary letter. We have the choices of  (a,b,c) of which there are  $2^{3} = 8$. Permutations result again in one of these 8 choices, so indeed there are no further arrangements.
In total, we have the number of size 3 subsets with pairwise distance of $k=2$ equals $2^{n-3} \cdot {n \choose 3} \cdot 2^{3} = 2^{n} \cdot {n \choose 3}$. 
The terms you have given match this formula.
As for other  $k$, the answer given here may be generalized, albeit with some more manual work. E.g. for $k=4$, 6 positions have to be looked at more closely, where the 3 words have to have letters  (a,b,c,d,e,f) , $(\bar a, \bar b, \bar c, \bar d, e, f)$ and $( a, b, \bar c, \bar d, \bar e, \bar f)$, however I expect that permutations play a role here such that for general $k$, we have that the number of size 3 subsets with pairwise distance of even k equals $ 2^{n} \cdot {{n} \choose {\frac{3 k}{2}}} \cdot $ (permutation factor). 
And it may be cumbersome to evaluate the permutation factor for larger k. 
