Suppose we look at all sets in ${[n] \choose k}$, for some $k \leq n/2$, and place them in layers according to the sum of their elements. Then, we say that two sets $A$ and $B$ from consecutive layers are connected by an edge, if there is some $i \in [n-1]$ for which $(i,i+1)A = B$. That is, if $A \Delta B = \{i,i+1\}$. Is it true that any two consecutive layers have a perfect matching, in the sense that all the elements of the layer with less sets are matched? For instance, let's take $n=6$, $k=3$. And let's match the layer with sum $12$ to the layer with sum $11$:


$(\{6,5,1\}, \{6,4,1\}), (\{6,4,2\},\{6,3,2\}), (\{5,4,3\}, \{5,4,2\})$.



Sums of sets in ${[n] \choose k}$
You are looking at two consecutive ranks of the poset denoted
$L(k,n-k)$. I proved the existence of a matching in
<http://math.mit.edu/~rstan/pubs/pubfiles/42.pdf>. A more elementary
proof based on linear algebra was later given by R. A. Proctor,
*Amer. Math. Monthly* **89** (1982), 721-734. Another elementary proof
based on linear algebra appears in my book *Algebraic Combinatorics*
(built into the proof of Corollary 6.10). It is an open problem to
find a combinatorial proof.


