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:
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|.
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:
- 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.
- Multiply. That is the whole calculation.
- 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|