Cartesian product of sets

A sum is odd only when one number is odd and the other is even. Set A has 2 odd and 1 even element, while set B has 1 odd and 2 even elements. Therefore, the number of favourable ordered pairs is (2 × 2) + (1 × 1) = 5. Hence, the correct answer is 5. Exam tip: For an odd sum, count odd-even and even-odd pairs separately and add them.

The possible pairs satisfying \(x+y=2\) are \((0,2)\), \((1,1)\), and \((2,0)\). Only \((0,2)\) and \((2,0)\) belong to \(A\times B\), because the first component must be from \(A\) and the second from \(B\). The pair \((1,1)\) is invalid since \(1\notin B\). Hence, there are 2 valid pairs. Exam tip: In a Cartesian product, check membership of each component in its corresponding set, while preserving the order.

The governing concept is the ordered structure of a Cartesian product together with maximisation of a sum. In A × B, the first coordinate must be an element of A and the second must be an element of B. To maximise x + y, choose the largest allowed first coordinate and the largest allowed second coordinate independently. The largest element of A is 4, and the largest element of B is 5, so the candidate pair is (4,5), whose sum is 4 + 5 = 9. Option B, (5,4), is not even in A × B because 5 is not in A. Options C and D are valid product pairs, but each has sum 7. Thus option A is the only correct answer.

From x−y=2, we get x=y+2. For y=1, x=3, giving the valid pair (3,1); for y=2, x=4, giving (4,2). Both x-values belong to A and both y-values belong to B, so there are 2 ordered pairs. Option B counts only (3,1) and misses (4,2). Exam tip: For each y∈B, calculate x=y+2 and check whether x belongs to A.

The governing idea is that A × B consists of ordered pairs with x ∈ A and y ∈ B, so both membership conditions and the equation xy = 8 must be satisfied. Check each possible x from A. For x = 1, y = 8, and (1, 8) is valid because 8 ∈ B. For x = 2, y = 4, giving (2, 4). For x = 3, y = 8/3, which is not in B. For x = 4, y = 2, giving (4, 2). Thus the complete solution set is {(1,8), (2,4), (4,2)}, containing three distinct ordered pairs. Since order is significant in a Cartesian product, the count is 3, so option A is correct.

Since (B-C={3,4}), each element of (A) pairs with (3) and (4). Find the difference first and then write the pairs.

The valid pairs are ((1,1),(1,2),(1,3),(2,2),(2,3),(3,3)). Include equal pairs when the condition is (\le).

In \(A\times B\), \(x\) is chosen from \(A\) and \(y\) from \(B\). The ordered pairs satisfying \(x+y>5\) are \((2,4),(3,3),(3,4)\), so there are 3 pairs. Pairs such as \((1,4)\) and \((2,3)\), for which \(x+y=5\), are excluded because the inequality is strict. Exam tip: carefully distinguish between \(>\) and \(\geq\).

If both sets are non-empty, at least one ordered pair is formed. Therefore (A\times B=\varnothing) cannot happen.

Order matters in Cartesian product, so it is generally not commutative. It may be equal in special cases such as (A=B) or because of an empty set.

(3\in A) is odd and (b\in B), so ((3,b)\in R). Apply the given condition to the Cartesian product.

Answer: A, A × B. The Cartesian product A × B is the set of all ordered pairs (a,b) for which the first entry belongs to A and the second entry belongs to B. In (1,3), 1 belongs to A and 3 belongs to B. In (2,4), 2 belongs to A and 4 belongs to B. Therefore both ordered pairs in R are members of A × B, so R ⊆ A × B. In B × A the first entry would have to come from B and the second from A; the given pairs do not follow that order. A ∩ B and A ∪ B are sets whose elements are individual numbers, whereas R contains ordered pairs, so they are not the suitable containing set here. Thus A is the only correct choice. Memory cue: in A × B, read the order as first from A, second from B; changing the order changes the product.

The governing idea is to apply the operation x + y to every ordered pair in the Cartesian product and then remove repeated results because the output is a set. First, A × B = {(0,2), (0,3), (1,2), (1,3)}. Their component sums are 0+2 = 2, 0+3 = 3, 1+2 = 3, and 1+3 = 4. Thus the resulting list is 2, 3, 3, 4. Sets do not record duplicate occurrences, so the second 3 is written only once. The required set is therefore {2,3,4}, making option A correct. Option B omits 4, option C lists original elements rather than sums, and option D is incomplete and incorrect.

All (3) elements of (B) are even and pair with (3) elements of (A). Hence (3\times3=9) pairs are obtained.

The odd elements in (A) are (1) and (3), and each pairs with (2) elements of (B). Therefore (2\times2=4) pairs are formed.

The direct answer is option A, (2,4). First find the intersection: A={1,2} and B={2,3} have only 2 in common, so A∩B={2}. Therefore (A∩B) × C = {2} × {3,4}. Its pairs are (2,3) and (2,4). Option A is correct because its first coordinate 2 belongs to A∩B and its second coordinate 4 belongs to C. Option B, (1,3), fails because 1 is not in A∩B, even though it is in A. Option C, (3,2), has the positions reversed: 3 is not in A∩B and 2 is not in C. Option D, (4,2), is also reversed and 4 is not in the first set. The safe method is intersection first, then form ordered pairs; never swap the two coordinates.

(B\cap C={5}), so (n(A\times(B\cap C))=3\times1=3). First find the intersection and then count the Cartesian product.

The governing conditions are Cartesian-product membership and the equation x + y = 5. The first coordinate must come from A and the second from B. Test each x in A: for x = 0, y = 5, and 5 belongs to B, so (0,5) is valid; for x = 2, y = 3, and (2,3) is valid; for x = 4, y = 1, and (4,1) is valid. These are all possible choices because A contains only 0, 2, and 4. Hence the solution set is {(0,5), (2,3), (4,1)}, which contains exactly three ordered pairs. Therefore option A is correct. The count is not based on the size of the entire product alone; the equation must also be checked.

The correct count is (4) for (y=1), (3) for (y=2), and (1) for (y=4). Thus total pairs are (4+3+1=8).

The governing relation is x/y = 2. Since every element of B is nonzero, we may safely multiply by y and rewrite the condition as x = 2y. Check each possible y from B: y = 1 gives x = 2, so (2,1) belongs to A × B; y = 2 gives x = 4, so (4,2) is valid; y = 4 gives x = 8, so (8,4) is valid. All three calculated x-values are in A, and B has no other elements. Thus the complete solution set is {(2,1), (4,2), (8,4)}, whose cardinality is 3. Therefore option A is correct. The other options result from omitting a valid pair or using an incorrect count.

In ((1,4)), (4\ne2\times1), so it is not in (R). Apply the given relation condition to each option.

The products are (1\cdot2=2), (1\cdot4=4), (1\cdot6=6), (3\cdot2=6), (3\cdot4=12), (3\cdot6=18). In a set, (6) is not written twice.

The direct answer is option B: 6 ordered pairs. In A × B, each element of A is paired with each element of B, and order matters. Set A has 3 elements: 1, 4, and 7. Set B has 2 elements: 0 and 2. Therefore n(A × B) = n(A)n(B) = 3·2 = 6. The pairs are (1,0), (1,2), (4,0), (4,2), (7,0), and (7,2), which confirms the count. Option A, 5, is one fewer than the correct total and has no valid counting basis. Option B, 6, correctly multiplies the two set sizes. Option C, 3, counts only the elements of A and ignores the choices from B. Option D, 2, counts only the elements of B and ignores A. Remember that ordered pairs are counted by choices for the first position multiplied by choices for the second position. Exam cue: never add the set sizes for a Cartesian product; multiply them.

In (A\times B), the first component comes from (A) and the second from (B). Each element of (A) pairs with all elements of (B).

Since (6\in A) and (4\in B), ((6,4)\in A\times B) is true. Check both positions separately for membership.