Cartesian product of sets

In the Cartesian product \(A\times B\), each element of \(A\) is paired with every element of \(B\). Therefore, \(n(A\times B)=n(A)\times n(B)=3\times2=6\), so the correct answer is 6. The values 5, 3 and 2 result from incorrect counting. Exam tip: For finite sets, use the formula \(n(A\times B)=n(A)n(B)\).

The Cartesian product A × B is defined as the set of all ordered pairs (x, y) such that x belongs to A and y belongs to B. The order is essential: the first coordinate must come from A, while the second must come from B. Since A has three elements and B has two, the product must contain 3 × 2 = 6 ordered pairs. Pairing each element of A with both elements of B gives (a,0), (a,1), (b,0), (b,1), (c,0) and (c,1). Thus option A is correct. Option B reverses the order and represents B × A, option C uses pairs from A only, and option D lists only three of the six required pairs.

Since (2\in A) and (3\in B), ((2,3)\in A\times B) is true. Check both positions separately in an ordered pair.

(B\cap C={4,6}), so (n(A\times(B\cap C))=4\times2=8). First find the intersection and then multiply.

The governing concepts are union cardinality and the product-cardinality rule. First form the union, remembering that a set records each distinct element only once: B ∪ C = {3,4} ∪ {4,5,6} = {3,4,5,6}. The common element 4 is therefore counted once, so n(B ∪ C) = 4. Also, n(A) = 2. For finite sets, n(X × Y) = n(X)n(Y), because every choice from X can be paired with every choice from Y. Hence n(A × (B ∪ C)) = 2 × 4 = 8, making option A correct. Option B results from counting the repeated 4 twice, while 6 and 12 do not follow the required multiplication after the union is formed.

Set difference A − B consists of elements that are in A but not in B. From A = {1, 2, 3, 4, 5}, remove 1, 3 and 5 because they belong to B; this leaves A − B = {2, 4}, which has two elements. The set C = {0, 2} also has two elements. For finite sets, the Cartesian product rule is n(X × Y) = n(X)n(Y). Therefore n((A − B) × C) = 2 × 2 = 4, so option A is correct. The elements themselves may be listed as (2,0), (2,2), (4,0) and (4,2). Options B, C and D result from failing to remove the elements of B or from using an incorrect counting rule.

(n(A\times B)=n(A)n(B)), so (35=5n(B)) and (n(B)=7). In such questions, use the formula in reverse too.

Both have (3\times4=12) elements. The number can be the same, but the order of pairs is generally different.

(A={1,2,3,4}) and (B={1,2}), so (n(A\times B)=4\times2=8). Convert set-builder form into roster form first.

The direct answer is option A, 8 pairs. First list the integers satisfying -2≤x≤1: A={-2,-1,0,1}, so n(A)=4. Set B={0,2}, so n(B)=2. For every one of the 4 choices from A, there are 2 choices from B, and each choice produces one ordered pair. Hence n(A × B)=n(A)n(B)=4×2=8. Explicitly, the pairs are (-2,0),(-2,2),(-1,0),(-1,2),(0,0),(0,2),(1,0),(1,2). Option A is correct. Option B, 6, results from missing one or more integers in A. Option C, 4, counts only the elements of A and ignores the second coordinate. Option D, 10, overcounts the product. Pay attention that the interval includes both endpoints -2 and 1 because the inequalities are ≤.

No first component can be taken from the empty set, so no ordered pair is formed. If either side is empty, the Cartesian product is (\varnothing).

(n(A\times A)=n(A)^2=2^2=4). Pairs with equal components such as ((p,p)) and ((q,q)) are also included.

The valid pairs are ((1,2),(1,4),(2,4),(3,4),(1,6),(2,6),(3,6),(4,6)), so the count is (8). List carefully to avoid overcounting.

For the condition x+y=4, check the possible values of x. When x=1, y=3; when x=2, y=2; and when x=3, y=1. All these pairs belong to A\times B. For x=0, y would have to be 4, but 4 is not in B. Thus the ordered pairs are (1,3), (2,2), and (3,1), giving a total of 3 pairs. Exam tip: determine the second coordinate from the condition and then verify that both coordinates belong to their respective sets.

For each element of \(A\), calculate \(y=2x\). This gives \(y=2\) for \(x=1\), \(y=4\) for \(x=2\), and \(y=8\) for \(x=4\). Thus, the valid pairs are \((1,2),(2,4),(4,8)\), giving a total of 3 pairs. A pair is valid only when \(x\in A\) and the corresponding \(y\in B\); merely counting the elements of either set is incorrect. Exam tip: for a condition of the form \(y=f(x)\), test every \(x\) from the first set and check whether the resulting \(y\) belongs to the second set.

The valid pairs are ((2,1),(2,2),(3,1),(3,3),(6,1),(6,2),(6,3)). In divisibility, note which component is the divisor.

When the first component (2) is fixed, the second component can be any of the (3) elements of (B). Hence there are (3) pairs.

When the second component (6) is fixed, the first component can be any of the (4) elements of (A). So there are (4) pairs.

(n(A\times B)=3\times2=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: {(1,4),(3,5)}. A relation from A to B is any subset of A × B. This means every ordered pair must have its first entry in A and its second entry in B. Here A = {1,2,3} and B = {4,5}. In option A, (1,4) and (3,5) both satisfy this rule, so the set is a valid relation. Option B reverses the order: (4,1) and (5,3) have first entries from B and second entries from A, so it is a subset of B × A, not A × B. Option C contains (1,6), but 6 is not in B. Option D contains 0, which is not in A, so its first pair is invalid; also the second pair uses 2 as a second component, not an element of B. A relation need not contain every element or every possible pair; it only needs to be a subset. Memory cue: for a relation from A to B, read each pair left to right as A then B.

(n(A\times B)=2\times3=6), so the number of subsets is (2^6). Use the formula (2^n) for counting subsets.

The governing concepts are Cartesian-product membership and intersection. In A × B, the first coordinate comes from A and the second from B, so A × B = {(1,2),(1,3),(2,2),(2,3)}. In B × A, the first coordinate comes from B and the second from A, giving B × A = {(2,1),(2,2),(3,1),(3,2)}. The intersection contains only ordered pairs that occur in both lists. The sole common pair is (2,2), because 2 belongs to both A and B. Thus the intersection has exactly one element and option A is correct. Pairs such as (1,2) and (2,1) may look related, but they are different ordered pairs because their coordinate order differs. This order distinction is essential.

In the Cartesian product \(A\times B\), the first coordinate comes from \(A\) and the second from \(B\). Thus, \(A\times B=\{(x,1):0\le x\le2\}\), so the value of \(y\) is always 1. Therefore, it is the horizontal line segment from \(x=0\) to \(x=2\) on \(y=1\). Option B interchanges the roles of the two coordinates. Exam tip: when the second set is a singleton \(\{c\}\), the product lies on the line \(y=c\).

By definition, A × B = {(a,b) : a ∈ A and b ∈ B}. The set A contains only -1, so every ordered pair in the product must have -1 as its first coordinate. The interval B = [2,5] contains every real number y satisfying 2 ≤ y ≤ 5, with both endpoints included. Consequently, A × B = {(-1,y) : 2 ≤ y ≤ 5}, which is option A. Geometrically, the product is the vertical line segment x = -1 between y = 2 and y = 5. Option B reverses the coordinate roles, option C incorrectly fixes the first coordinate at 2, and option D includes only one pair rather than all pairs generated by the interval. Thus A is the only complete description.

(n(B\times C)=1\times2=2), so (n(A\times(B\times C))=2\times2=4). Count the inner Cartesian product first.