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Let $S$ be a set of consisting of $10$ elements. The number of tuples of the form $(A,B)$ such that $A$ and $B$ are subsets of $S$, and $A \subseteq B$ is ___________
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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 $A\subseteq B$:-

$A$ $B$
0 0
0 1
1 1

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.

 

 

Answer:

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