A question says: "If |A| = 7 and |B| = 9, find |A \times B|." A student who has not internalised the pattern reaches for pen and paper, starts listing pairs, and stops three minutes later with the right answer. A student who has internalised the pattern writes 63 and moves on.

The rule is as simple as it looks:

|A \times B| = |A| \cdot |B|

No exceptions, no asterisks, no "only if the sets are disjoint." Whenever you see "cardinality of a Cartesian product," your first (and usually only) move is to multiply.

The recognition cue

Any of the following phrasings triggers the rule:

  • "Find |A \times B|"
  • "How many ordered pairs are there in A \times B?"
  • "The number of elements in A \times B is ___"
  • "A has m elements, B has n elements. Find |A \times B|."
  • "Find |A \times A|" (same rule — this is |A|^2)
  • "R \subseteq A \times B, find the maximum size of R"

The last one is the sneakiest — it does not say "Cartesian product" explicitly, but any relation lives inside A \times B, so the maximum is exactly |A \times B| = |A| \cdot |B|.

Why the reflex always works

For every element a \in A, you can pair it with any of the |B| elements in B. That gives |B| pairs with first coordinate a. Repeat for each of the |A| elements of A, and you get |A| groups of |B| pairs each. The total is |A| \cdot |B|.

Grid showing 4 times 5 equals 20 cells for a Cartesian productA grid with 4 rows labelled a1 through a4 on the left and 5 columns labelled b1 through b5 along the top. Each of the 20 cells contains a red dot representing an ordered pair. The product 4 times 5 equals 20 is shown at the bottom. A B → b₁ b₂ b₃ b₄ b₅ a₁ a₂ a₃ a₄ |A| = 4, |B| = 5, |A × B| = 4 × 5 = 20
Every row contributes $|B| = 5$ pairs; there are $|A| = 4$ rows; total is $4 \times 5 = 20$ cells. This multiplication principle is what the Cartesian product cardinality rule formalises.

Why: a cell in the grid is labelled by a pair (a, b), and the grid is a rectangle with |A| rows and |B| columns. Area of a rectangle equals length times width. Counting pairs is the same kind of multiplication.

Small worked examples

Example 1. A = \{1, 2, 3\}, B = \{x, y\}. Find |A \times B|.

|A| = 3, |B| = 2, so |A \times B| = 3 \cdot 2 = 6. (Verification by listing: \{(1,x),(1,y),(2,x),(2,y),(3,x),(3,y)\} — six pairs. Matches.)

Example 2. |A| = 11, |B| = 13. Find |A \times B| and |B \times A|.

|A \times B| = 11 \cdot 13 = 143. |B \times A| = 13 \cdot 11 = 143. The cardinality is the same, even though the two sets of pairs are different (one has pairs starting with elements of A, the other starts with B).

Example 3. A = \{1, 2, 3, 4, 5, 6\}. Find |A \times A|.

|A \times A| = 6 \cdot 6 = 36. This counts the number of ordered pairs from A to A — exactly the sample space when you roll two distinguishable dice.

Example 4. A relation R is a subset of A \times B where |A| = 5 and |B| = 4. What is the maximum possible size of R?

Maximum relation is the full Cartesian product: |R|_{\max} = |A| \cdot |B| = 20.

Example 5 (triple product). |A| = 2, |B| = 3, |C| = 4. Find |A \times B \times C|.

The rule extends: |A \times B \times C| = |A| \cdot |B| \cdot |C| = 2 \cdot 3 \cdot 4 = 24.

The reflex, step by step

When you see a cardinality-of-Cartesian-product question:

  1. Identify the sets and their sizes. Sometimes |A| and |B| are given as numbers; sometimes you have to count them from an explicit listing first.
  2. Multiply. That is the whole calculation.
  3. If the question asks for |A \times A|, you compute |A|^2. If for A \times B \times C, multiply all three. The pattern does not change.

Do not reach for a Venn diagram, do not enumerate pairs, do not split into cases. Cartesian product is one of the rare operations where the naive count happens to be the right count — no inclusion-exclusion correction, no overlap to subtract.

Why it is different from |A \cup B|

A common confusion: students apply inclusion-exclusion (|A \cup B| = |A| + |B| - |A \cap B|) to the Cartesian product. Do not. Cartesian product is about pairs, not elements of a single combined set. There is no overlap to correct for because A \times B contains ordered pairs, which cannot equal elements of A or B individually.

So:

  • |A \cup B| = |A| + |B| - |A \cap B| (union — add, then subtract overlap)
  • |A \times B| = |A| \cdot |B| (product — multiply)

Two different operations, two different formulas. The cue word in the question tells you which. "Union" → add and subtract. "Cartesian product" or "ordered pairs" → multiply.

One exam anti-pattern to avoid

The question sometimes tries to confuse you:

"Let |A| = 3, |B| = 4, |A \cap B| = 1. Find |A \times B|."

A student in a panic subtracts the intersection. Do not. The intersection is a red herring — |A \times B| depends only on |A| and |B|, not on how A and B overlap as sets of elements. The answer is still 3 \times 4 = 12.

The only time |A \cap B| matters for a Cartesian product question is if the problem asks for something like "how many pairs (a, b) \in A \times B with a = b" — and even that is a filtered subset, not the full product. The count of the full product is always the plain multiplication.

The exam reflex

  • See "|A \times B|" or "number of ordered pairs" or "size of the relation set."
  • Identify |A| and |B| from the problem.
  • Write |A| \cdot |B| and compute. Done.

Once this reflex is wired, you will never spend more than a few seconds on this class of question. The time savings compound — in a 180-minute JEE paper, every two-second problem you bank gives you more time for the seven-minute problems that actually need thinking.

Related: Set Operations · Relations · Cartesian Product Grid · Given |A|, |B|, |A ∪ B|