Cartesian product of sets

Since (B) is non-empty, every element of (A) appears as a first component. Therefore the set of first components is (A).

Second components always come from (B). Since (A) is non-empty, every element of (B) appears as a second component.

The sum is even when both components have the same parity. Here such pairs are (2\times2+2\times1=6).

Using x+y=7, we get y=5 for x=2, y=3 for x=4, and y=1 for x=6. Thus the valid ordered pairs are (2,5), (4,3), and (6,1), giving a total of 3. For x=0, y would be 7, but 7 is not in B. Exam tip: calculate y=7−x for each x and then check whether y belongs to B.

The governing idea is that an ordered pair (a, b) in A × B uses a from A and b from B, and its component sum is a + b. To maximize this sum, choose the largest available first component from A and the largest available second component from B. The largest element of A is 8, and the largest element of B is 7, so the pair (8, 7) has sum 8 + 7 = 15. This is option C. The sums of the other valid choices are 7 + 8 = 15 for option A, 8 + 4 = 12 for option B, and 5 + 7 = 12 for option D; however, option A is not even in A × B because 7 is not in A and 8 is not in B. Thus C is the unique valid pair with greatest sum.

From x−y=2, we get x=y+2. For each value of y in B, the valid pairs are (3,1), (4,2), and (5,3). All three first components belong to A, so there are 3 ordered pairs. Choosing 4 is incorrect because the number of elements in A alone does not determine the number of valid pairs. Exam tip: substitute each value from the smaller set and check whether the resulting value belongs to the other set.

The governing concepts are Cartesian-product membership and ordered-pair counting. We need x from A and y from B such that xy = 6. Checking the positive factor pairs of 6 using the given set gives (1, 6), because 1 × 6 = 6; (2, 3), because 2 × 3 = 6; (3, 2), because 3 × 2 = 6; and (6, 1), because 6 × 1 = 6. All four pairs belong to A × B. Ordered pairs preserve position, so (2, 3) and (3, 2) are different pairs even though their products are equal. No other listed values produce 6 with a partner from the set. Therefore option C, 4, is correct. A count of 2 would incorrectly treat reversed pairs as identical.

(B-C={3,9}), so each element of (A) pairs with (3) and (9). Find the difference first and then write the pairs.

Cartesian product is a set of ordered pairs. In the correct form, the first component is from (A) and the second is from (B).

For (y=2), there are (2) pairs; for (y=3), (3) pairs; and for (y=4), (4) pairs. Total pairs are (2+3+4=9).

For x=1, the sums with elements of B are 3, 5, and 7, so none qualifies. For x=2, only (2,6) has a sum greater than 7. For x=5, the pairs (5,4) and (5,6) have sums 9 and 11. Thus, there are 3 valid ordered pairs. Exam tip: a strict inequality such as x+y>7 excludes pairs whose sum is exactly 7.

The Cartesian product A × B consists of all ordered pairs whose first component is selected from A and whose second component is selected from B. If A is empty, there is no possible choice for the first component. This remains true even when B is nonempty, because every pair requires one first component from A.

Here A = ∅, so no ordered pair can be formed at all. Consequently, A × B = ∅, regardless of which nonempty set B is. The condition that B is nonempty does not create a pair, because it only provides possible second components. Therefore the statement is true and option B is correct. This is also consistent with the counting rule: the product has 0 × n(B) = 0 elements.

A relation consisting of ordered pairs is commonly viewed as a subset of a Cartesian product. In A × B, the first entry must belong to A and the second entry must belong to B. For (0, 3), we have 0 ∈ A and 3 ∈ B. For (2, 4), we have 2 ∈ A and 4 ∈ B. Hence both members of R are in A × B, so R ⊆ A × B. The reverse product B × A would require first components from B and second components from A, which does not fit these pairs. A ∩ B and A ∪ B are sets of individual elements rather than sets of ordered pairs, so they cannot be the appropriate containing Cartesian product here. Therefore option C is correct.

The ordered pairs in \(A \times B\) are \((1,2),(1,5),(2,2),(2,5)\). Their component sums are \(3,6,4,7\), so the set of sums is \(\{3,4,6,7\}\). Option D includes only \(3\) and \(7\), omitting \(4\) and \(6\). Exam tip: In a Cartesian product, pair every element of the first set with every element of the second set.

All (3) elements of (B) are even and pair with (4) elements of (A). Hence (4\times3=12) pairs are formed.

The prime elements in (A) are (2,3,5), and each pairs with (2) elements of (B). Therefore (3\times2=6) pairs are formed.

(A\cap B={2,3}), so the first component must be (2) or (3) and the second must be from (C). ((3,0)) satisfies this condition.

Use the cardinality rule |X × Y| = |X| |Y|, but first simplify the second factor. The union B ∪ C collects every element appearing in either set: B ∪ C = {2, 3, 4, 5}. Removing all elements of A = {1, 2, 3} leaves (B ∪ C) − A = {4, 5}, which has two elements. Since A has three elements, the Cartesian product A × {4,5} contains one pair for each of the 3 choices from A and each of the 2 choices from the second set. Therefore its size is 3 × 2 = 6. Option A ignores one second-component choice, option B confuses the union’s size with the product size, and option D would incorrectly multiply 3 by 3.

The relation R is formed by selecting from A × B only those ordered pairs whose coordinates satisfy x + y = 7. Since A has 1, 2, and 3 and B has 4 and 5, check each possible first coordinate with the available second coordinates. For x = 1, the required y would be 6, which is not in B. For x = 2, y = 5, giving (2,5). For x = 3, y = 4, giving (3,4). Both pairs belong to A × B and satisfy the equation, so R = {(2,5),(3,4)}. Option B includes pairs summing to 6, option C omits one valid pair, and option D includes (1,4), whose sum is 5. Therefore option A is correct.

First find the intersection, because only elements common to both B and C can be used in the second coordinate. B = {2,3,4} and C = {3,4,5}, so B ∩ C = {3,4}, which has cardinality 2. Set A has cardinality 3. Applying the Cartesian-product rule gives |A × (B ∩ C)| = |A| × |B ∩ C| = 3 × 2 = 6. Equivalently, each of the three elements of A can be paired with either 3 or 4, producing six ordered pairs. Option A forgets one possible second coordinate, option C would use three elements in the intersection, and option D incorrectly multiplies by four. Thus option B is the unique correct answer.

The set difference B − C contains elements that are in B but not in C. Since B = {1,2} and C = {2,3}, the element 2 is removed because it belongs to C, while 1 remains because it does not. Thus B − C = {1}. Now form A × {1}. Each element of A, namely 0 and 1, is paired with the only available second component 1. Therefore A × (B − C) = {(0,1),(1,1)}. Option B uses the removed element 2, option C treats the operation as if B and C were both available without subtraction, and option D would be correct only if B − C were empty. Hence option A correctly applies both set difference and the ordered-pair rule.

The Cartesian product A × B contains every ordered pair with its first coordinate selected from A and its second coordinate selected from B. Since each set has four elements, the total number of pairs is |A × B| = 4 × 4 = 16. The condition a ≠ b excludes exactly the diagonal pairs in which both coordinates are equal: (1,1), (2,2), (3,3), and (4,4). There are four such excluded pairs. Therefore the required number is 16 − 4 = 12, making option C correct. Option A counts only the pairs that violate the condition. Option D includes those forbidden diagonal pairs, and option B has no valid counting basis. This is an application of complementary counting within a Cartesian product.

Answer: C, 6. First calculate the intersection, because the expression uses A ∩ B before forming the Cartesian product. The common elements of A = {1,2,3} and B = {2,3,4} are 2 and 3, so A ∩ B = {2,3} and |A ∩ B| = 2. Set C = {1,3,5} has three elements, so |C| = 3. For finite sets X and Y, the Cartesian product contains one ordered pair for every choice of an element of X and an element of Y; therefore |X × Y| = |X|·|Y|. Hence |(A ∩ B) × C| = 2·3 = 6. A may result from counting only the intersection incorrectly. B has no valid product calculation behind it. D incorrectly uses |A|·|C| = 3·3 and fails to form the intersection first. Memory cue: intersection first, then multiply the two cardinalities.

Interpret the set-builder conditions carefully. Assuming N denotes the positive natural numbers, A contains the integers from 2 through 6 inclusive: A = {2,3,4,5,6}, so |A| = 5. B contains positive odd natural numbers less than 8: B = {1,3,5,7}, so |B| = 4. The Cartesian-product rule states that each of the 5 choices from A can be paired with each of the 4 choices from B. Therefore |A × B| = |A| |B| = 5 × 4 = 20. Option B is correct. The other choices arise from omitting a member, treating the interval incorrectly, or multiplying the wrong cardinalities.

The governing idea is that A × B contains ordered pairs (a,b) with a from A and b from B, while the stated condition filters those pairs. Test each possible first component from A. For a = 1, b = 2, and 2 belongs to B, giving (1,2). For a = 2, b = 4, giving (2,4). For a = 3, b = 6, but 6 is not in B, so no pair is formed. For a = 4, b = 8, giving (4,8). Therefore there are actually three valid pairs: (1,2), (2,4), and (4,8). Thus option C is mathematically correct; the supplied key B is incorrect and must be revised.