Number of chains in a symmetric chain decomposition

Question

I need to show that the number of chains of length $n-2k$ in a symmetric chain decomposition of Boolean Lattice $B_n$ is $\binom{n}{k}-\binom{n}{k-1}$. But I have no idea how to do it. I also have a question: is there always a symmetric chain decomposition containing a given symmetric chain in $B_n$? Any help of you would be very much appreciated!

Answer 1

First, count the number of chains whose length is at least $n-2k$. A chain is at least $n-2k$ in length if and only if it contains a subset at level $k$. There are exactly $\binom{n}k$ subsets at level $k$, and exactly one chain through each, so $$ \#\{\text{chains with length $\ge n-2k$}\}=\binom{n}k $$ Then, $$ \#\{\text{chains with length $n-2k$}\}=\quad\#\{\text{chains with length $\ge n-2k$}\}\\\hspace{7.4cm}-\#\{\text{chains with length $\ge n-2(k-1)$}\} $$

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For the second question, take an arbitrary symmetric chain decomposition, and then apply an appropriate permutation of the elements to each chain to make sure the desired chain is included.