Cartesian product of sets

Among pairs with sum (6), only ((1,5)) and ((2,4)) have (a<b). Apply combined conditions one by one.

The correct answer is option C, 9. In an ordered pair (a,b) from A×B, b must be divisible by a, meaning b/a is an integer. For a=2, the elements of B divisible by 2 are 4, 6, 8 and 12: four pairs. For a=3, the suitable elements are 6, 9 and 12: three pairs. For a=4, the suitable elements are 4, 8 and 12: two pairs. The total is 4+3+2=9. Option A, 7, misses two valid pairs. Option B, 8, also misses one valid pair. Option C is correct because all nine valid ordered pairs have been counted. Option D, 10, counts one extra pair, probably by treating a number that is not a multiple of a as valid. Notice that order matters: the first entry must come from A and the second from B. The safest method is to test each a separately against every element of B.

For (a=1,2,3,4), the counts of (b) are (6,4,2,2), totaling (14). In product conditions, the limit changes for each (a).

Direct answer: Option C, 12. The size of a Cartesian product is found by multiplying the sizes of the sets. A has 3 elements and B has 2, so \\(|A\\times B|=3\\cdot2=6\\). The set \\( (A\\times B)\\times C \\) has 6 choices for its first component and 2 choices for its second component, because C has 2 elements. Hence its size is \\(6\\cdot2=12\\). Option A, 7, incorrectly adds the set sizes. Option B, 10, uses another incorrect addition. Option C is correct because it applies multiplication twice. Option D, 24, would require four times as many choices and is not supported by the given sets. Even though the objects are nested ordered pairs, each choice is still counted by multiplication. Memory cue: product size equals the product of set sizes.

There are (16) total pairs and (4) diagonal pairs, so (16-4=12). Removing the diagonal leaves pairs with (a\ne b).

a + b is divisible by 2 precisely when a and b have the same parity. The even elements are 2 in A and 2 in B, giving 2 × 2 = 4 pairs. The odd elements are 3 in A and 2 in B, giving 3 × 2 = 6 pairs. Therefore, the total number of pairs is 4 + 6 = 10. Exam tip: count even-even and odd-odd pairs separately for divisibility by 2.

The pairs with sum divisible by (4) are ((1,3),(2,2),(3,1),(4,4),(5,3)). In such questions, count using remainders.

For (a=1,2,3,4), the counts of (b) are (4,3,2,1), so there are (10) pairs. In inequalities, check the boundary for each first component.

In (A\times B), the first component comes from (A) and the second from (B). Changing positions is a common mistake.

Set \(A\) contains \(3,4,5,6,7,8\), so \(|A|=6\). Since \(\mathbb{N}\) is specified as the set of positive integers, \(B=\{1,2,3,4\}\), giving \(|B|=4\). For Cartesian products, \(|A\times B|=|A|\times|B|=6\times4=24\). Hence, the correct answer is 24. Exam tip: count the elements of both sets first, then multiply their cardinalities.

The correct answer is option C: 12. The size of a Cartesian product of finite sets is found by multiplying the sizes of all sets. Here |A|=2 because A has 1 and 2; |B|=3 because B has 3,4,5; and |C|=2 because C has 6,7. Therefore |A×B×C|=|A||B||C|=2×3×2=12. Every ordered triple chooses one element from A, one from B, and one from C. Option A, 7, does not follow the multiplication rule. Option B, 10, is also not the product of the three set sizes. Option C, 12, is correct. Option D, 18, is an overcount and may come from an incorrect operation. The order of positions is fixed, but each independent choice multiplies the number of possibilities. Memory cue: for a product, multiply the number of choices at each position.

In a Cartesian product, each element of the first set is paired with every element of the second set. Hence, \(|B\times A|=|B|\times|A|=4\times2=8\). Choosing 16 usually results from incorrectly using \(4\times4\), even though the second set has only 2 elements. Exam tip: use \(|X\times Y|=|X||Y|\); reversing the order may change the pairs, but not their total number.

From \(a+b=9\), we get \(b=9-a\). For \(a=1,2,3,4,5\), the corresponding values of \(b\) are \(8,7,6,5,4\), respectively. Only \((3,6)\) and \((5,4)\) have the first component in \(A\) and the second component in \(B\). Hence, the number of required ordered pairs is 2. Exam tip: For each \(a\in A\), calculate \(b=9-a\) and check whether \(b\in B\).

For (a=0,1,2,3,4), the counts of (b) are (4,4,3,2,1), totaling (14). Count row-wise in inequality questions.

The governing concept is the definition of a Cartesian product together with systematic counting under an inequality. An ordered pair (a,b) must use a from A and b from B. Fix b first. If b = 1, the condition becomes a ≥ 3, so a can be 3, 4, 5, or 6: four pairs. If b = 2, it becomes a ≥ 5, so only a = 5 and 6 work: two pairs. If b = 3, it becomes a ≥ 7, and no element of A satisfies it. Therefore the total is 4 + 2 + 0 = 6, making option B correct. The other values arise from including an invalid boundary value, reversing the inequality, or counting elements without respecting the ordered-pair condition.

For every (a\in A), (a^2\in B), so (5) pairs are formed. Check each first component in a rule-based relation.

The governing concept is divisibility within a Cartesian product. The statement a divides b means b/a is an integer, so we inspect every a in A and count eligible b-values from B. For a = 2, the values 6, 8, and 12 are divisible by 2, giving three pairs. For a = 3, the values 6, 9, and 12 are divisible by 3, giving three more pairs. For a = 4, the values 8 and 12 are divisible by 4, giving two pairs. Thus the total is 3 + 3 + 2 = 8, so option C is correct. The value 7 usually results from forgetting that 4 divides 12; smaller distractors omit one or more other valid pairs. Each pair remains ordered, with a from A and b from B.

The suitable pairs are ((2,2),(2,4),(2,6),(4,2),(8,2)), so the count is (5). Check common factors carefully in (\gcd) questions.

Since (|A\times B|=20), the number of ways to choose exactly (2) pairs is (\binom{20}{2}=190). Use combinations for exact-size subsets.

A relation is a subset of (A\times B), and (|A\times B|=15), so the number is (2^{15}=32768). Number of relations uses a power of (2).

First find the union: \(B\cup C=\{2,4,6,8\}\), which has 4 elements. Set \(A\) also has 4 elements. For a Cartesian product, \(|A\times D|=|A|\cdot|D|\), so \(|A\times(B\cup C)|=4\times4=16\). As an exam tip, count repeated elements in a union only once before multiplying cardinalities.

First find the intersection: \(B\cap C=\{2,4\}\), so \(|B\cap C|=2\). For a Cartesian product, \(|A\times(B\cap C)|=|A|\cdot|B\cap C|=3\cdot2=6\). Therefore, the correct answer is 6. Choosing 9 incorrectly treats the intersection as having three elements. Exam tip: determine the cardinality of the resulting set before applying \(|X\times Y|=|X||Y|\).

The governing concepts are set difference and the Cartesian product. Set difference B − C retains elements that belong to B but do not belong to C. Since B = {2,3,4,5} and C = {3,5,7}, removing the common elements 3 and 5 leaves B − C = {2,4}. Now form A × {2,4}. Every element of A must be paired with each element of {2,4}, with the A-element written first. This gives (1,2), (1,4), (3,2), (3,4), (5,2), and (5,4), six ordered pairs in all. Therefore option A is correct. Options B and C incorrectly retain elements removed by the difference, while option D wrongly assumes that the difference is empty.

((A\times B)\cap(A\times C)=A\times(B\cap C)), and (B\cap C={3}). Hence the cardinality is (3\cdot1=3).

((A\times B)\cup(A\times C)=A\times(B\cup C)), and (B\cup C={2,3,4,5}). Thus there are (2\cdot4=8) elements.