If $A = \{1, 2\}$ and $B = \{3, 4, 5\}$, find $A \times B$ and $n(A \times B)$.

If A = \1, 2\ and B = \3, 4, 5\, find A × B and n(A × B).

Example · Example 2

A = \{1, 2\}

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16% · 5/31 Questions

✓ Free question

Step 1: $A\times B=\{(a,b):a\in A,\,b\in B\}$ , so pair every element of $A=\{1,2\}$ with every element of $B=\{3,4,5\}$ .

Step 2: Taking $a=1$ : $(1,3),(1,4),(1,5)$ . Taking $a=2$ : $(2,3),(2,4),(2,5)$ .

Step 3: $n(A\times B)=n(A)\times n(B)=2\times3=6$ , matching the $6$ pairs listed.

✓Final answer

A\times B=\{(1,3),(1,4),(1,5),(2,3),(2,4),(2,5)\}

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CA Foundation 2026Set jan-20261 markMCQ
$A=\{2,\ 3\}$
›Reveal solutionSolution

$B\cap C=\{5\}$ , so $A\times\{5\}=\{(2,5),(3,5)\}$ .

Step 1 — Intersection $B\cap C$

Keep only the elements common to both sets:
$B\cap C=\{4,5\}\cap\{5,6\}=\{5\}.$
Step 2 — Cartesian product $A\times(B\cap C)$

Pair each element of $A=\{2,3\}$ with each element of $\{5\}$ , writing each ordered pair as (element from $A$ , element from $B\cap C$ ):
$A\times\{5\}=\{(2,5),\ (3,5)\}.$
Watch out

Order matters in an ordered pair. The first coordinate must come from $A$ and the second from $B\cap C$ — a pair like $(5,2)$ (option C) reverses this and is wrong. Also compute the intersection before the product, not after.

Tip

When one of the sets in a Cartesian product has just a single element, the product simply "attaches" that element to each member of the other set — no need to enumerate a full grid.

✓Final answer
$\{(2,\ 5),\ (3,\ 5)\}$
CA Foundation 2025Set jan-20251 markMCQ
$A = \{a, b, p\}$
›Reveal solutionSolution

$n(A \cup C) = 6$ , $n(B) = 2$ , so $n[(A \cup C) \times B] = 6 \times 2 = 12$ .

Step 1 — Form the union
$A \cup C = \{a, b, p\} \cup \{p, q, r, s\} = \{a, b, p, q, r, s\}$
The element $p$ is common and counted only once, so $n(A \cup C) = 6$ .

Step 2 — Note the size of $B$
$B = \{2, 3\} \;\Rightarrow\; n(B) = 2$
Step 3 — Cardinality of the Cartesian product
$n(X \times Y) = n(X) \times n(Y) = 6 \times 2 = 12$
Why the other options are wrong: (D) 16 counts $p$ twice ( $7 \times 2$ -ish slip) or mis-sizes a set; (A) 8 uses $n(A)=4$ ?; (B) 20 mishandles the union.

Watch out

In the union, count the shared element $p$ only ONCE — $n(A\cup C) = 6$ , not 7.

Tip

$n(A \cup C) = n(A) + n(C) - n(A \cap C) = 3 + 4 - 1 = 6$ if you prefer the inclusion–exclusion route.

✓Final answer
$(C) 12$
CA Foundation 2023Set jun-20231 markMCQ
$A=\{a, b, c\}, B=\{b, c, d\}$
›Reveal solutionSolution

A − B = {a}, B ∩ C = {c, d}, so {a} × {c, d} = {(a, c), (a, d)}.

Step 1 — Set difference

A = {a, b, c}, B = {b, c, d}: elements of A not in B ⇒ A − B = {a}.

Step 2 — Intersection

B = {b, c, d}, C = {a, d, c}: common elements ⇒ B ∩ C = {c, d}.

Step 3 — Cartesian product
$\{a\} \times \{c, d\} = \{(a, c), (a, d)\}$
Watch out

Ordered pairs: the first coordinate must come from A − B, so (c, a) and (d, a) are wrong.

Tip

|A − B| × |B ∩ C| = 1 × 2 = 2 pairs — a quick count of how many pairs to expect.

✓Final answer
$\{(a, c), (a, d)\}$

Note

This answer is verified — independently worked and reviewed by our experienced subject lecturers before we published it.

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Q.If $A = \{1, 2\}$ and $B = \{3, 4, 5\}$ , find $A \times B$ and $n(A \times B)$ .

Concept understanding — Cartesian Product of Sets

Step 1 — Intersection $B\cap C$

Step 2 — Cartesian product $A\times(B\cap C)$

Step 1 — Form the union

Step 2 — Note the size of $B$

Step 3 — Cardinality of the Cartesian product

Step 1 — Set difference

Step 2 — Intersection

Step 3 — Cartesian product

Q.If A={1,2}A = \{1, 2\}A={1,2} and B={3,4,5}B = \{3, 4, 5\}B={3,4,5}, find A×BA \times BA×B and n(A×B)n(A \times B)n(A×B).

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Q.If $A = \{1, 2\}$ and $B = \{3, 4, 5\}$ , find $A \times B$ and $n(A \times B)$ .

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