If A = a, b, c, B = x, y find A×B, B×A, A×A, B×B.

Exercise 5.2 · Q58

Q. If A = {a, b, c}, B = {x, y} find A\timesB, B\timesA, A\timesA, B\timesB.

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✓ Free question

$A\times B$ pairs each of A's 3 elements with each of B's 2 elements (3×2=6 pairs): $\{(a,x),(a,y),(b,x),(b,y),(c,x),(c,y)\}$ . $B\times A$ reverses the roles (2×3=6 pairs): $\{(x,a),(x,b),(x,c),(y,a),(y,b),(y,c)\}$ — note this is a different set of pairs from $A\times B$ since order matters. $A\times A$ pairs A with itself (3×3=9 pairs): $\{(a,a),(a,b),(a,c),(b,a),(b,b),(b,c),(c,a),(c,b),(c,c)\}$ . $B\times B$ pairs B with itself (2×2=4 pairs): $\{(x,x),(x,y),(y,x),(y,y)\}$ .

✓Final answer

A\times B = {(a,x),(a,y),(b,x),(b,y),(c,x),(c,y)}, ; B\times A = {(x,a),(x,b),(x,c),(y,a),(y,b),(y,c)}, ; A\times A = {(a,a),(a,b),(a,c),(b,a),(b,b),(b,c),(c,a),(c,b),(c,c)}, ; B\times B = {(x,x),(x,y),(y,x),(y,y)}

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 = {a, b, c}, B = {x, y} find A×B, B×A, A×A, B×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 = {a, b, c}, B = {x, y} find A×B, B×A, A×A, B×B.

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