Cartesian product of sets

Set \(P\) has 3 elements and set \(Q\) has 2 elements. Therefore, the number of ordered pairs in the Cartesian product is \(n(P\times Q)=n(P)\times n(Q)=3\times2=6\). Hence, the correct answer is 6. Exam tip: For the Cartesian product of two finite sets, multiply their cardinalities rather than adding them.

(8) stays in the first position and elements of (B) come in the second position. Order does not change even with a singleton set.

In ((7,8)), (7\in X) and (8\in Y), so it is correct. In membership, check the correct position of both components.

In (A\times B), the second component always comes from (B). In exams, match the second position with the second set.

There is no element in (R) for the first component, so no pair is formed. Cartesian product with an empty set is empty.

For two finite sets, the number of ordered pairs in their Cartesian product is \(n(A\times B)=n(A)\times n(B)\). Thus, \(7\times 3=21\), so the correct answer is 21. The value 10 results from addition and is therefore incorrect. Exam tip: In \(A\times B\), the first component is chosen from A and the second from B, so the numbers are multiplied.

(B) has (3) elements and (A) has (2) elements, so (3\times 2=6). Reversing order changes the pairs, not the count.

The definition uses the ordered pair ((x,y)), where (x\in A) and (y\in B). This formula is useful in every listing question.

The governing concept is equality of ordered pairs. Two ordered pairs are equal only when their corresponding components are equal in the same positions. Compare the first components of (a, 12) and (9, b): a = 9. Then compare the second components: 12 = b, so b = 12. Therefore option B is correct. Option A reverses the positions and incorrectly assigns 12 to a and 9 to b. Option C correctly identifies a but wrongly makes b equal to 9, while option D reverses the required first-component comparison. The order of components matters in an ordered pair.

(A) has (2) elements and (B) has (4) elements, so (2\times 4=8). Counting elements first is the safest method.

The order of elements in a set is not important, so (A=B). Remember that order in a set and order in an ordered pair are different ideas.

In ((6,3)), the first component is (6), but (6\notin A). To identify the wrong option, start by checking the first component.

The direct answer is option B: 4. For finite sets, the number of ordered pairs in a Cartesian product is n(A×B)=n(A)×n(B). The question gives n(A×B)=24 and n(A)=6. Substitute these values: 24=6×n(B). To find n(B), divide both sides by 6: n(B)=24/6=4. Option B is correct. Option A, 3, would give 6×3=18 pairs, not 24. Option C, 6, would give 6×6=36 pairs, so it also fails the given total. Option D, 18, would give 6×18=108 pairs and is far too large. The important idea is that the product count is shared equally among the choices from A: each of the 6 elements of A must pair with every element of B. Therefore 24 total pairs divided by 6 partners per A-element gives 4 elements in B. Memory cue: unknown factor = product count ÷ known factor.

(B) is empty, so no second component is available and no pair is formed. Since count is asked, write (0).

The governing rule is the cardinality formula for a Cartesian product: if a finite set X has m elements and a finite set Y has n elements, then |X × Y| = mn. Here C = {2, 3, 5}, so |C| = 3. Since both factors are C, the number of ordered pairs is |C × C| = |C| × |C| = 3 × 3 = 9. The pairs are (2,2), (2,3), (2,5), (3,2), (3,3), (3,5), (5,2), (5,3), and (5,5). Order matters, so (2,3) and (3,2) are different pairs. Therefore option C is correct. Option A counts only one choice, option B uses an incorrect addition, and option D has no basis in the product rule.

(A) is not empty and still the product is empty, so (B) must be empty. In an empty product, at least one set is empty.

The governing definition is A × B = {(a, b) : a belongs to A and b belongs to B}. In the product {9, 10} × {4}, the first component may be 9 or 10 because it comes from the first set. However, the second component must come from the singleton second set {4}. Thus the only possible ordered pairs are (9,4) and (10,4), and both have y = 4. Therefore, option C is correct. Options A and B are possible values of x, not y. Option D incorrectly transfers the alternatives for the first component to the second component. Respecting the order of the factors and the positions in the ordered pair resolves the question immediately.

The governing definition is membership in a Cartesian product. An ordered pair (x,y) belongs to A × B precisely when x ∈ A and y ∈ B. In this question the first factor is the singleton set {12}. Therefore the first component can have only one possible value: x = 12. The possible pairs are (12,1), (12,2), and (12,3), and each confirms that the first component remains 12. The numbers 1, 2, and 3 come from the second factor, so they can be possible values of y, not x. The order of factors is essential; reversing the product would change the component positions. Hence option A is the unique correct answer.

The first component (4) is fixed and the second component comes from the (2) elements of (B). With a fixed first component, the count is (n(B)).

The second component (2) is fixed and the first component can be any of the (2) elements of (A). With a fixed second component, the count is (n(A)).

Both have count (3\times 1=3), but the order of components is reversed. Equal count does not mean equal sets.

Changing positions changes an ordered pair. Ordered pairs are equal only when corresponding components are equal.

The governing definition of a Cartesian product is (u, v) ∈ A × B if and only if u ∈ A and v ∈ B. For the ordered pair (5,5), the first coordinate is checked against A and the second coordinate is checked against B. Since 5 is an element of A = {4,5} and also an element of B = {5,6}, both membership conditions are satisfied. Therefore (5,5) belongs to A × B, so option A is correct. The equality of the two coordinates causes no problem: Cartesian-product pairs do not have to contain different numbers. Option B incorrectly imposes such a restriction, option C contradicts the given set A, and option D is false because A is not empty.

The number of ordered triples in a Cartesian product equals the product of the cardinalities of the individual sets. Here, \(n(A)=2\), \(n(B)=2\), and \(n(C)=1\), so \(n(A\times B\times C)=2\times2\times1=4\). Therefore, option B is correct. Exam tip: multiply the number of elements in each set; do not add them.

In ((1,0)), (1\notin A) and (0\notin B), so it is not in (A\times B). Check membership of both components.