Cartesian product of sets

For three finite sets, the cardinality of their Cartesian product is n(A × B × C) = n(A) × n(B) × n(C). Set A contains 2 elements, set B contains 1 element, and set C contains 3 elements. Therefore, n(A × B × C) = 2 × 1 × 3 = 6. Each ordered triple chooses its first component from A, its second component from B, and its third component from C. The six triples are (1,3,4), (1,3,5), (1,3,6), (2,3,4), (2,3,5), and (2,3,6). Hence option C is correct. Options A and B undercount the independent choices, while option D does not follow the product-of-cardinalities rule.

An ordered pair (a, b) belongs to A × B exactly when a ∈ A and b ∈ B. For (0,0), the first 0 is in A and the second 0 is in B, so it is included. For (1,2), 1 ∈ A and 2 ∈ B, so it is included. For (1,0), both membership conditions also hold, so it is included. However, in (2,1), the first component is 2, and 2 is not an element of A; also, the second component 1 is not in B. Therefore (2,1) cannot belong to A × B, making option D correct. The distractors are valid because they respect the required first-set and second-set positions.

Elements of (A) in the second component are also in (B), so (C\times A\subset C\times B). With subsets, check the position of the related component.

In the first product, (1) is in the first position, and in the second, (2) is in the first position. Order matters even with one element each.

Each ordered pair can be treated like a point, and there are (3\times 2=6) pairs. The same multiplication rule applies in coordinate questions.

There are (2) colors and (3) sizes, so (2\times 3=6) choices are formed. Real-life combinations are counted using Cartesian product.

(n(B\times A)=n(B)n(A)=1\times 3=3). Reversing order does not change the count, but it changes the order of pairs.

The first component (3) is fixed, and the second component is (4) or (5) from (B). With a fixed first component, only elements of the second set vary.

The second component (2) is fixed, and the first component is (5) or (7) from (A). In such pairs, only the first position changes.

(A\times B) has (2\times 2=4) elements, and they are ordered pairs. Do not forget the pair form in the answer.

The defining condition for membership in a Cartesian product is (a, b) ∈ A × B if and only if a ∈ A and b ∈ B. In the ordered pair (2, 4), the first component is 2. Since 2 is an element of A = {1, 2}, the first condition is satisfied. The second component is 4, and 4 is an element of B = {3, 4, 5}, so the second condition is also satisfied. Therefore (2, 4) belongs to A × B, making option A correct. Option B reverses the required set membership and is false; 2 is not in B and 4 is not in A. Option C is false because 2 and 4 are unequal, and option D is false because A is not empty.

The answer is option B, 4. The set-builder description says A contains natural numbers not greater than 2. Using the usual school convention N={1,2,3,...}, A={1,2}, so n(A)=2. Set B={10,20} has n(B)=2. The Cartesian-product rule gives n(A×B)=n(A)n(B)=2×2=4. Option A counts only one set, while C and D do not follow the multiplication rule. The actual pairs are (1,10),(1,20),(2,10),(2,20), confirming four. If a text defines natural numbers to include 0, A would have three elements, so the convention should be checked; under the supplied school convention, B is correct.

The equal-component pairs are ((1,1),(2,2),(3,3)). In such questions, do not count all (9) pairs; count only the required condition.

Elements of (A) come in the first position and (1) from (B) comes in the second position. A negative element is used like any ordinary element.

The first component (\text{Mon}) is from (A) and the second component (\text{evening}) is from (B). Therefore, the pair belongs to (A\times B).

The second component (4) is fixed and the first component comes from (1) or (2) in (A). With a second-component condition, the first set varies.

A Cartesian product lists ordered pairs. The first member must come from set A, while the second member must come from set B. Since A contains 3 and 4 and B contains only 5, every possible pair has 5 in its second position. Thus the product contains exactly (3,5) and (4,5), with no repeated or unordered entries.

To check the options, pair each element of A with the only element of B: 3 gives (3,5), and 4 gives (4,5). Therefore option A is correct. Option B reverses the order and represents B 5c7ftimes7f A, option C is an ordinary collection rather than ordered pairs, and option D is false because neither set is empty.

The smallest element of (A) is (1), and it pairs with the (3) elements of (B). For a fixed first component, the answer is (n(B)).

The largest element of (B) is (2), and the (3) elements of (A) can be first components with it. For a fixed second component, the answer is (n(A)).

The pairs are ((1,3)) and ((2,2)), and both have sum (4). In condition-based questions, first think of all possible pairs.

The direct answer is option C: 4. Set A contains 1 and 3, both odd numbers. Set B contains 2 and 4, both even numbers. An odd number plus an even number is always odd. Therefore every ordered pair in A×B has an odd sum. The complete product is (1,2), (1,4), (3,2), and (3,4), so there are 2×2=4 pairs. Option C is correct. Option A, 0, is wrong because none of the pairs has an even sum; all four sums are odd. Option B, 2, would count only part of the possible pairs and misses two valid pairs. Option D, 6, is impossible because A×B has only 4 pairs in total. The first component must be selected from A and the second from B, giving four combinations. A useful cue is parity: odd plus even always gives odd, so count all product pairs.

The Cartesian product \\(A\\times B\\) contains every ordered pair \\((a,b)\\) with the first component selected from \\(A\\) and the second selected from \\(B\\). Here, \\(A=\\{2,4\\}\\) and \\(B=\\{1,3\\}\\). Both possible first components, 2 and 4, are even. An even number multiplied by any integer is even, so every pair in the Cartesian product has an even product.

There are \\(2\\) choices for the first component and \\(2\\) choices for the second component. By the multiplication principle, the total number of ordered pairs is \\(2\\times2=4\\). They are \\((2,1)\\), \\((2,3)\\), \\((4,1)\\), and \\((4,3)\\), and all products are even. Hence option C is correct.

The equal-component pairs are ((1,1)) and ((2,2)). In such questions, common elements of both sets are useful.

The pairs are ((1,2)) and ((2,3)), where the difference between the second and first component is (1). In condition-based questions, first list possible pairs and then check.

In (A\times B), the first component comes from (A) and the second from (B). The full list must have (2\times 2=4) ordered pairs.