Suppose $A_1, A_2, \ldots, A_{30}$ are thirty sets each having 5 elements and $B_1, B_2, \ldots, B_n$ are $n$ sets each with 3 elements, let $\bigcup_{i=1}^{30} A_i = \bigcup_{j=1}^{n} B_j = S$ and each element of $S$ belongs to exactly 10 of the $A_i$'s and exactly 9 of the $B_j$'s. Then $n$ is equal to (A) $15$ (B) $3$ (C) $45$ (D) $35$

Suppose A₁, A₂, …, A₃₀ are thirty sets each having 5 elements and B₁, B₂, …, Bₙ are n sets each with 3 element

NCERT Exemplar · Q29

A_1, A_2, \ldots, A_{30}

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Concept understanding — Cartesian Product

Cartesian Product: From Intuition to Definition

Imagine you're ordering a pizza. You have two choices to make: the size (Small, Medium, Large) and the topping (Cheese, Pepperoni, Veggie). How many different pizzas can you order?

You can pair each size with each topping:

Small + Cheese, Small + Pepperoni, Small + Veggie
Medium + Cheese, Medium + Pepperoni, Medium + Veggie
Large + Cheese, Large + Pepperoni, Large + Veggie

That's $3 \times 3 = 9$ possible pizzas. What you just did — systematically pairing every element of one set with every element of another — is the Cartesian product in action.

The Intuition

The Cartesian product is a way to combine two sets to create a new set of ordered pairs. The order matters: (Small, Cheese) is different from (Cheese, Small) — one is a pizza order, the other is nonsense.

Think of it like a multiplication table for sets. If set $A$ has $m$ items and set $B$ has $n$ items, their Cartesian product has $m \times n$ items.

Note

The name comes from René Descartes, who used this idea to create the coordinate plane — every point $(x, y)$ on a graph is an element of the Cartesian product of the x-axis and y-axis.

The Precise Definition

Let $A$ and $B$ be two sets. The Cartesian product of $A$ and $B$ , written $A \times B$ , is the set of all ordered pairs $(a, b)$ where $a$ is from $A$ and $b$ is from $B$ .

A \times B = \{(a, b) \mid a \in A \text{ and } b \in B\}

The vertical bar means "such that." So read it as: "The set of all ordered pairs (a, b) such that a belongs to A and b belongs to B."

Key Properties to Remember

Order matters: $A \times B$ is generally not the same as $B \times A$ . For example, if $A = \{1, 2\}$ and $B = \{x, y\}$ :
- $A \times B = \{(1, x), (1, y), (2, x), (2, y)\}$
- $B \times A = \{(x, 1), (x, 2), (y, 1), (y, 2)\}$
These are different sets because the pairs are ordered differently.
Size formula: If $|A| = m$ and $|B| = n$ , then $|A \times B| = m \times n$ . This holds even if one set is empty — then the product is empty.
Empty set: $A \times \emptyset = \emptyset$ and $\emptyset \times B = \emptyset$ . You can't form any pairs if one set has nothing to contribute.

Watch out

A common mistake: thinking $A \times B$ contains all possible combinations of elements from $A$ and $B$ without caring about order. But $(a, b)$ and $(b, a)$ are different pairs unless $a = b$ . Always treat ordered pairs as distinct based on position.

Examples to Cement the Idea

Example 1: $A = \{1, 2\}$ , $B = \{3, 4\}$

A \times B = \{(1, 3), (1, 4), (2, 3), (2, 4)\}

Four pairs, as expected ( $2 \times 2 = 4$ ).

Example 2: $A = \{a\}$ , $B = \{1, 2, 3\}$

A \times B = \{(a, 1), (a, 2), (a, 3)\}

Three pairs — every element of $B$ gets paired with the single element of $A$ .

Example 3: $A = \{0, 1\}$ , $B = \{0, 1\}$

A \times B = \{(0, 0), (0, 1), (1, 0), (1, 1)\}

This is the set of all possible 2-bit binary strings — a foundation for computer science.

Why This Matters

The Cartesian product is the mathematical backbone of: …

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Count the total element-memberships in two ways: through the $A_i$ 's and through the $B_j$ 's. Since each element of $S$ appears in exactly 10 of the $A_i$ 's and exactly 9 of the $B_j$ 's, equating these counts gives $n = \boxed{45}$ .

The heart of this problem is a double-counting argument. We have the same universal set $S$ covered by two different families of sets, and we're told precisely how many times each element appears in each family. By counting the total number of (element, set) pairs in two different ways, we can extract the unknown $n$ .

Let $|S| = s$ denote the number of elements in the universal set.

Counting through the $A_i$ family:

Each of the 30 sets $A_i$ contains exactly 5 elements. If we sum up all elements across all $A_i$ 's (with repetition), we get:

\sum_{i=1}^{30} |A_i| = 30 \times 5 = 150

But this counts each element of $S$ multiple times—specifically, each element is counted once for every $A_i$ it belongs to. Since each element of $S$ belongs to exactly 10 of the $A_i$ 's, we have:

\sum_{i=1}^{30} |A_i| = \sum_{x \in S} (\text{number of } A_i \text{'s containing } x) = s \times 10

Therefore:

s \times 10 = 150 \implies s = 15

Counting through the $B_j$ family:

Similarly, each of the $n$ sets $B_j$ contains exactly 3 elements. Summing across all $B_j$ 's:

\sum_{j=1}^{n} |B_j| = n \times 3

Each element of $S$ belongs to exactly 9 of the $B_j$ 's, so: …

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