There is a set $X=\{1,2,...,n\}$ and its subset $A$. Where $|A|=k$. Lets choose $l\le 2^{n}$
How many are there ways of choosing $l$ different subsets of $X$ that its common intersection is equal to $B$?
I tried to translate it into language of chains (theory of partially ordered sets) and find characteristic function but so far i can't see the solution.
First, note that every one of the $l$ subsets must contain $B$, so we can remove $B$ from each subset without an issue. Suppose $B=\{m+1,...,n\}$. So, let $C=\{1,...,m\}$.
Now, our goal is to find $l$ subsets of $C$ with empty intersection. That's equivalent to finding $l$ subsets of $C$ with their union equal to $C$, since we can then take complements of each subset.
To find how this is possible, we first partition $C$ into $l$ subsets. This will ensure that the union is $C$. We can do this via stars and bars to get $m+l-1\choose l-1$. From here, you have to do some careful counting, but you should be able to find the desired quantity.
