Let A and B two sets containing 2 elements and 4 elements respectively. The number of subsets of A ×

Q: Wait, why is the total number of elements in $A \times B$ just $n(A) \times n(B)$?

Think of the Cartesian product as a grid. If you have $2$ rows (elements of $A$) and $4$ columns (elements of $B$), every intersection point is a unique ordered pair $(a, b)$. Since each row has $4$ points, $2$ rows give you $2 \times 4 = 8$ points. It is the fundamental counting principle in action!

Q: Why do we use complementary counting instead of just adding up the subsets with $3, 4, 5, 6, 7,$ and $8$ elements?

You absolutely could! But calculating $\binom{8}{3} + \binom{8}{4} + \binom{8}{5} + \binom{8}{6} + \binom{8}{7} + \binom{8}{8}$ is much more prone to calculation errors and takes significantly longer. Complementary counting is a powerful tool in your JEE arsenal—it turns a long, tedious addition problem into a quick subtraction.

Q: Is the empty set really a subset of $A \times B$?

Yes, absolutely! By definition, the empty set $\emptyset$ is a subset of every set. When we calculate the total number of subsets as $2^n$, we are including the empty set (where $r=0$) and the set itself (where $r=n$). Never forget to account for it!

Question

Let A and B two sets containing 2 elements and 4 elements respectively. The number of subsets of $A 	imes B$ having 3 or more elements is

Options:

(A) 256

(B) 220

(C) 219

(D) 211

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future IITian! Today, we are going to peel back the layers of a seemingly simple problem in set theory. It is not just about numbers; it is about understanding the structure of the universe we are working with.

Let us start by visualizing our two sets, $A$ and $B$. We are told that set $A$ has exactly $2$ elements, and set $B$ has $4$ elements.

Imagine plotting set $B$ along the horizontal axis and set $A$ along the vertical axis. This coordinate-like setup helps us see how these two sets interact when we combine them.

### Constructing the Cartesian Product

Now, let us construct the Cartesian product, $A 	imes B$. This product consists of all possible ordered pairs where the first element comes from set $A$ and the second from set $B$.

Mathematically, the number of elements in $A 	imes B$ is simply the product of the number of elements in $A$ and $B$. That is:

$$n(A 	imes B) = n(A) 	imes n(B) = 2 	imes 4 = 8$$

Each of these eight intersection points represents a unique ordered pair in our Cartesian product. These eight points form the entire universe of elements from which we will construct our subsets.

### The Power of the Power Set

Now, how many total subsets can we form from these eight elements? Recall the fundamental rule of set theory: if a set has $n$ elements, the total number of possible subsets is $2^n$.

Since our Cartesian product has $n = 8$ elements, the total number of subsets is:

$$2^8 = 256$$

This includes everything from the empty set to the full set of eight elements.

### The Complementary Strategy

The question asks for the number of subsets having $3$ or more elements. If we try to calculate this directly, we would have to find the number of subsets with $3, 4, 5, 6, 7,$ and $8$ elements, and then add them all up.

That sounds like a lot of tedious work! Instead, let us use a clever shortcut called **complementary counting**.

We can find the total number of subsets and subtract those that have fewer than $3$ elements—meaning subsets with $0, 1,$ or $2$ elements.

### The Final Calculation

Let us calculate these excluded cases one by one. First, how many subsets have $0$ elements? There is only one such subset, which is the empty set, represented by:

$$\binom{8}{0} = 1$$

Next, how many subsets have exactly $1$ element? These are called singleton subsets, and since we have $8$ elements to choose from, we have:

$$\binom{8}{1} = 8$$

Finally, let us find the number of subsets with exactly $2$ elements. This is given by the combination formula:

$$\binom{8}{2} = \frac{8 	imes 7}{2 	imes 1} = 28$$

Adding all our excluded cases together, we get $1 + 8 + 28 = 37$ excluded subsets. We are now ready for the final step!

To find the number of subsets with $3$ or more elements, we simply subtract our $37$ excluded subsets from the total of $256$ subsets:

$$256 - 37 = 219$$

We have successfully solved the problem using a highly efficient complementary counting approach. The final answer is **219**.

Remember, in JEE, the most elegant path is often the one that saves you time and minimizes errors. Keep practicing, and keep falling in love with the logic behind the math!

Animated Solution for Mathematics - Sets and Relations: Let A and B two sets containing 2 elements and 4 elements respectively. The number of subsets of $A \times B$ having 3 or more elements is

Select Answer:

Visualized Solution

Visualizing Sets $A$ and $B$

Constructing the Cartesian Product $A \times B$

Ploting the 8 Ordered Pairs

Total Number of Subsets $2^{n}$

The Complementary Counting Strategy

Subsets with $0$ and $1$ Element

Subsets with Exactly $2$ Elements

Final Subtraction

The Sigma Insight: Cartesian Product of Sets

Analyzing the Setup

Constructing the Cartesian Product

The Power of the Power Set

The Complementary Strategy

The Final Calculation

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Comprehension Passage

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