Cartesian product of sets

In ((2,5)), the first element is not from (A) and the second is not from (B). Reversing order is a common Cartesian product mistake.

The direct answer is option C, 3 elements. First calculate the intersection B∩C. B={2,3,4} and C={1,3,5} have only 3 in common, so B∩C={3}; its size is 1. Set A={1,2,3} has size 3. The Cartesian product A × (B∩C) therefore has n(A)n(B∩C)=3×1=3 elements. They are (1,3), (2,3), and (3,3). Option A, 6, would incorrectly use two elements in the intersection. Option B, 9, would multiply 3×3 and wrongly treat B∩C as having three elements. Option C is correct because the intersection has exactly one element. Option D, 12, overcounts and has no correct product basis. The reliable order is: find the intersection first, count it, then multiply by the size of A.

(B\cup C={2,4,6,8}), so (n(A\times(B\cup C))=2\times4=8). Do not count common elements twice in a union.

The governing rules are set difference and the cardinality formula n(X × Y) = n(X)n(Y) for finite sets. First remove from A every element that also belongs to B. Since B = {4, 8}, we obtain A − B = {2, 6}, which has two elements. Set C = {1, 3, 5} has three elements. Therefore n((A − B) × C) = n(A − B) × n(C) = 2 × 3 = 6, so option B is correct. The ordered pairs would be (2,1), (2,3), (2,5), (6,1), (6,3), and (6,5). Option A may result from an incorrect count, option C ignores the product structure, and option D incorrectly uses all four elements of A instead of removing B.

The direct answer is option C, 8. The Cartesian product A × B contains every ordered pair whose first member comes from A and second member comes from B. If A has 6 members and B has n(B) members, each of the 6 choices from A can be matched with every choice from B. Therefore, n(A × B) = n(A)n(B). Substituting the given values gives 48 = 6n(B). Dividing both sides by 6 gives n(B) = 8. Option A, 6, is only the size of A, not B. Option B, 42, comes from subtracting 6 from 48, but Cartesian-product counting uses multiplication, not subtraction. Option C, 8, follows exactly from the formula and calculation. Option D, 54, comes from adding 6 to 48 and has no valid basis. Remember: when one set size and the product size are known, divide the product size by the known set size.

Both have (4\times5=20) elements. The number can be the same but the order of pairs is generally different.

Here, \(A=\{2,3,4,5\}\), so \(n(A)=4\), and \(B=\{1,2,3\}\), so \(n(B)=3\). For a Cartesian product, \(n(A\times B)=n(A)\times n(B)\); therefore, \(4\times3=12\). Hence, option C is correct. Option D would result only if 0 were included in \(\mathbb{N}\) and consequently in \(B\); in this question, natural numbers are explicitly taken as positive integers. Exam tip: The number of ordered pairs in \(A\times B\) is always the product of the cardinalities of the two sets.

The second component must come from (B), but (B=\varnothing). Hence no ordered pair is formed and the product is (\varnothing).

Set A has 3 elements. Therefore, the number of ordered pairs in A × A is n(A × A) = n(A) × n(A) = 3 × 3 = 9. Pairs with equal components, such as (m, m), (n, n), and (o, o), are also included, so 6 is incorrect. Exam tip: For any set A, n(A × A) = [n(A)]².

The condition \(x=y\) means that the common elements of both sets must be used. Since \(A\cap B=\{3,4\}\), the valid ordered pairs are \((3,3)\) and \((4,4)\). Therefore, there are 2 pairs. Exam tip: the number of pairs in \(A\times B\) with equal coordinates is \(|A\cap B|\).

The valid pairs are ((0,1),(0,3),(1,3),(2,3),(0,5),(1,5),(2,5),(3,5)), so the count is (8). Careful listing prevents overcounting.

In A × B, x must come from A and y must come from B. To satisfy x + y = 5, test the possible x-values from A and find y = 5 − x. A candidate counts only when this y is also an element of B. This prevents counting numerical solutions that do not form valid Cartesian-product pairs.

For x = 1, y = 4, giving (1,4). For x = 2, y = 3, giving (2,3). For x = 4, y = 1, giving (4,1). For x = 0, y would be 5, but 5 is not in B. Hence there are exactly three valid pairs, so option C is correct. Each pair has both the required sum and the required set membership.

Check y=2x for every element of A: x=1 gives y=2, x=3 gives y=6, and x=5 gives y=10. All three resulting values belong to B, so the valid ordered pairs are (1,2), (3,6), and (5,10), giving a total of 3. Counting only two pairs overlooks the valid pair obtained from x=5. In an exam, compute y for each element of A and then verify that the result lies in B.

The valid pairs are ((4,1),(4,2),(4,4),(6,1),(6,2),(8,1),(8,2),(8,4)). In divisibility, note the positions of divisor and dividend carefully.

The governing concept is a relation formed from a Cartesian product. An ordered pair (x, y) is eligible only when x belongs to A, y belongs to B, and the equation y = x² is satisfied. Test every element of A: x = 1 gives y = 1² = 1, so (1, 1) is valid; x = 2 gives y = 2² = 4, so (2, 4) is valid; and x = 4 gives y = 4² = 16, so (4, 16) is valid. All three calculated second components lie in B. Hence exactly three ordered pairs satisfy the condition, and option B is correct. Option A omits one valid pair, while options C and D count pairs that either fail the equation or are not generated by the stated rule.

When the first component (4) is fixed, the second component can be any of the (4) elements of (B). Hence (4) pairs are formed.

When the second component (20) is fixed, the first component can be any of the (4) elements of (A). So there are (4) pairs.

(n(A\times B)=2\times3=6), and every relation is a subset of (A\times B). Therefore the total number of relations is (2^6).

The direct answer is option A: {(2,1),(6,3)}. A relation from A to B must be a subset of A × B, so the first member of every pair must come from A = {2,4,6}, and the second must come from B = {1,3}. In option A, (2,1) and (6,3) satisfy both requirements, so it is valid. Option B reverses the direction: its first entries 1 and 3 come from B, not A, so it belongs to B × A. Option C contains (4,5); although 4 is in A, 5 is not in B. Option D contains 0 as a first entry, but 0 is not in A, so it fails immediately; its second pair also uses 2, which is not in B. A relation may contain only some pairs, so it does not need to use every element of A or B. The important test is membership and order for each pair. Memory cue: source set A comes first, target set B comes second.

The direct answer is option A, 2^6. First count the elements of the Cartesian product. Set A has 3 elements and set B has 2 elements, so n(A × B) = 3 × 2 = 6. A set with n elements has 2^n subsets because each element has two independent choices: it is either included or not included. Thus A × B has 2^6 subsets, which equals 64. Option A is correct. Option B, 6, is only the number of elements in A × B, not the number of its subsets. Option C, 2^5, would be used for a five-element set, but this product has six elements. Option D, 3^2, uses the sizes of A and B in the wrong way; it is not the subset formula. The safe method is: multiply the set sizes first, then raise 2 to that product. Do not confuse elements with subsets.

The common pairs are ((2,2),(2,4),(4,2),(4,4)). In intersection, the whole ordered pair must belong to both sets.

Both have (2\times3=6) elements, but changing the order changes pairs. Cartesian product is generally not commutative.

In the Cartesian product \(A\times B\), the first coordinate comes from \(A\) and the second coordinate comes from \(B\). Thus, \(A\times B=\{(x,3):2\le x\le5\}\), which is a horizontal line segment on \(y=3\). Option A incorrectly interchanges the roles of the coordinates. Exam tip: when the second set is a singleton \(\{c\}\), the Cartesian product lies on the line \(y=c\).

The governing definition is A × B = {(a, b) : a ∈ A and b ∈ B}. Since A contains only the number 2, the first coordinate of every ordered pair must be 2. The second coordinate can be any real number y in the closed interval [−1, 4]. Hence A × B = {(2, y) : −1 ≤ y ≤ 4}, which is option B. Geometrically, these points form the vertical line segment x = 2 from y = −1 to y = 4, including both endpoints because the interval is closed. Option A reverses the coordinate order. Option C includes only the two endpoints and omits all interior values, while option D contains only one pair rather than the complete product.

Set \(A\) has 3 elements, \(B\) has 2, and \(C\) has 1. Thus, \(A\times B\) contains \(3\times2=6\) ordered pairs. Each of these pairs is combined with the only element of \(C\), so \((A\times B)\times C\) has \(6\times1=6\) elements. Choosing 12 incorrectly treats \(C\) as if it had more than one element. Exam tip: use \(n(X\times Y)=n(X)\times n(Y)\).