Answer:- $3^{10}$
We’ll be using binary indicators to solve this problem.
Let 0- indicate the presence in the set; 1- indicate absence from the set.
We’ll be using a tuple of 2 indicator variables corresponding to 2 sets (A,B) for each element in S, each indicator will tell us whether that element is present or absent in set A & B respectively.
Listing down are valid combinations of the indicators satisfying the given constraint $A\subseteq B$:-
Note:-$(A,B)=(1,0)$ is an invalid combination as it’s not possible for an element to be present in A & not in B.
Now, it’s evident that there’re 3 choices for each element in $S$, making the ans $3^{|S|}\ or\ 3^{10}$
Let’s try to generalize it to$(A_{1},A_{2},....A_{k}),where\ A_{i}\subseteq A_{i+1}\ \forall i, i<k, and\ A_{k}\subseteq S$, $|S|=n$
Initially all indicators are set to 0, or the indicator tuple looks like $(0,0,…….k\ times)$
$i=1$, If we set $A_{k}=1$ we get a valid tuple, similarly
$i=2$, if $A_{k-1}\&\&A_{k}==1$, or both $A_{k}\&A_{k-1}$ are set, we get a valid tuple, and so on:-
In $i^{th}$ step we’re setting last i indicators, till all the indicators are 1 when i=k
As this process iterates $k$ times, we get $k$ valid combinations, another valid combination is our initial setup when all k indicators are 0, or the element of $S$ is present in none of the subsets $A_{i}$.
So, in total we’ve $(k+1)$ valid bit combinations for every element in $S$.
Or, $(k+1)^{|S|}=(k+1)^{n}$ valid tuples of the given form.
