How to understand a fact in the proof of Sperner's Theorem? Consider the maximal chain in $(S,\varsubsetneqq)$. A maximal chain is made up of the null set, a subset of size 1, a subset of size 2,..., and finally the set itself. There are n choises for the subset of size 1. The subset of size 2 must contain the subset of size 1 that was picked previously. Thus there are n-1 choices for the second element in the subset. Continuing to build up the maximal chain in this way, we see that there are n! maximal chains. Suppose that A is an antichain. If $s\in A$ and $|s|=k$, then $s$ belongs to $k!(n-k)!$ maximal chains.
I don't understand why the last sentence is ture, can anyone help me to understand the last sentence of this proof? Thanks very much!!
 A: You didn't say what is $S$, and I do not see what is the relevance of $A$. 
But, take $|s|=k$. A maximal chain containing $s$ would contain one set 
of cardinality $k-1$, one set of cardinality $k-2$, etc. Also, going in the other direction, it will contain one set of cardinality $k+1$, one set of cardinality $k+2$, 
etc (which of course needs to be made more precise, e.g. if $k=n$ then there will be no $k+1$, etc, but it ought to be clear anyway). 
So starting from $k$ we may go down (picking sets of smaller and smaller size), or up (picking sets of bigger and bigger size) to form a maximal chain. 
Going down is the same as scratching out (someone posted an answer but let me finish anyway) elements from $s$, one after the other. First we have $k$ many choices to scratch out an element, then $k-1$ choices, etc. This clearly leads to the $k!$ term in the product $k!(n-k)!$. Similarly, going up, is the same as picking elements from 
the remaining $n-k$ many elements that are not in $s$ and adding them to $s$. First we have $n-k$ many choices, next $n-k-1$ many choices, etc, which accounts for the term $(n-k)!$ in the product $k!(n-k)!$
A: Each subset of size $k$ is an element of the same number of maximal chains, and two different subsets of size $k$ are never elements of the same maximal chain. There are $n\choose k$ different subsets of size $k$, and each is therefore an element of the same fraction, $\frac{1}{n \choose k}$, of the maximal chains. Since there are $n!$ maximal chains, each subsets of size $k$ is an element of $\frac{1}{n \choose k}\cdot n!=k!(n-k)!$ maximal chains. (The assumption that $s$ is an element of a particular antichain is irrelevant.)
