Animated Solution for Mathematics - Sets and Relations: Let $A=\{1,2,3,4,5,6,7\}$ and $B=\{3,6,7,9\}$. Then the number of elements in the set $\{C\subseteq A:C\cap B\ne \varnothing \}$ is ________

Imagine you are standing in front of a massive library of subsets. You have a set $A=\{1,2,3,4,5,6,7\}$ and a set $B=\{3,6,7,9\}$.

Your task is to find how many subsets $C$ of $A$ have at least one element in common with $B$. This is the classic 'at least one' problem that haunts many JEE aspirants.

If you try to count them directly—subsets with one common element, two common elements, three common elements—you will quickly find yourself drowning in a sea of cases. But wait! There is a more elegant path.

First, let us define our universe. We are forming subsets $C$ from $A$.

Since $A$ has $7$ elements, the total number of possible subsets is:

This is our total pool. Now, we need to filter this pool. We want subsets where $C \cap B eq \emptyset$.

This means $C$ must contain at least one of the elements from the intersection $A\cap B$. Let us find that intersection: $A\cap B=\{3,6,7\}$. There are $3$ elements here.

This is where the magic happens. Instead of counting the 'good' subsets (those that intersect with $B$), let us count the 'bad' ones (those that do not intersect with $B$ at all).

A subset is 'bad' if it contains absolutely no elements from $A\cap B$. This means the subset $C$ must be formed entirely from the elements in $A$ that are not in $B$.

Let us calculate $A\setminus B=\{1,2,4,5\}$. There are $4$ such elements. The number of subsets we can form using only these $4$ elements is:

These $16$ subsets are the ones that completely avoid $B$.

Now, we simply subtract the 'bad' subsets from the total. We have $128$ total subsets and $16$ subsets that are disjoint from $B$.

The number of valid subsets is:

It is that simple! By shifting our perspective from what we want to what we do not want, we have turned a complex counting problem into a simple subtraction.

Remember this for your exam: whenever you see 'at least one' or 'non-empty intersection', the complement method is your best friend. It is fast, it is reliable, and it is the mark of a true problem-solver. The final answer is 112.