Cartesian product of sets

Correct answer: B. (6)

Explanation: The direct answer is option B: 6 pairs. A Cartesian product contains ordered pairs \((a,b)\), with the first coordinate from A and the second from B. Since both sets contain 1 through 5, list sums divisible by 4. Possible sums range from 2 to 10, so the relevant sums are 4 and 8. For sum 4, the pairs are \((1,3),(2,2),(3,1)\): 3 pairs. For sum 8, the pairs are \((3,5),(4,4),(5,3)\): 3 more pairs. Total \(3+3=6\). Option B is correct. Option A, 5, misses one valid ordered pair. Option C, 7, counts an extra pair whose sum is not divisible by 4. Option D, 8, is also too large; it may result from counting pairs without checking the set limits. Ordered pairs matter, so \((1,3)\) and \((3,1)\) are separate, although both are counted here. Memory cue: list multiples of 4 in the possible sum range, then count ordered pairs for each sum.

Correct answer: B. (7)

Explanation: The number of ordered pairs with sum (5) or (10) is (7). Complementary remainders give a quick count.

Correct answer: C. (12)

Explanation: There are (16) pairs with (a\mid b), and pairs where (b\mid a) also holds are ((1,1),(2,2),(3,3),(4,4)). Hence (16-4=12).

Correct answer: B. (5)

Explanation: For each (x), (R) maps it to (6-x) and then back to (x). Thus (R\circ R={(x,x):x\in A}) has (5) pairs.

Correct answer: C. (7)

Explanation: Both numbers must be multiples of (2), and after division the reduced pair must be coprime, giving (7) pairs. Factor out the common divisor in (\gcd) questions.

Correct answer: C. (10)

Explanation: The direct answer is option C: 10 pairs. The complement of R relative to \(A\times B\) contains all pairs that are not in R. Since R is defined by \(a+b\ge6\), its complement is defined by \(a+b<6\). For \(a=1\), b can be 1, 2, 3, or 4: 4 pairs. For \(a=2\), b can be 1, 2, or 3: 3 pairs. For \(a=3\), b can be 1 or 2: 2 pairs. For \(a=4\), only b=1 works: 1 pair. Thus the total is \(4+3+2+1=10\). Option C is correct. Option A, 8, omits two valid pairs. Option B, 9, omits one valid pair. Option D, 11, counts one invalid pair, usually a pair with sum 6 or more. The complement is taken inside \(A\times B\), not among all possible integers. Memory cue: change \(\ge6\) to its opposite \(<6\), then count row by row.

Correct answer: C. 9

Explanation: For \(a+b\) to be even, \(a\) and \(b\) must have the same parity. For \(ab\) to be divisible by \(3\), at least one of \(a\) or \(b\) must be \(3\), since \(3\) is the only multiple of \(3\) in the set. Because \(3\) is odd, both entries must be odd. Among the odd values \(\{1,3,5\}\), the ordered pairs with \(a=3\) or \(b=3\) number \(3+3-1=5\), where \((3,3)\) is counted only once. Thus the correct answer is \(5\), which is not present among the supplied options. The original options contain a factual error. Exam tip: apply the parity condition first, then filter the remaining ordered pairs using divisibility.

Correct answer: A. (0)

Explanation: (R^{-1}) has the condition (a>b), which cannot hold together with (a<b). Therefore the intersection is empty.

Correct answer: A. 4

Explanation: For \(c=1\), the condition becomes \(a+b=3\). The ordered pairs from \(A\) and \(B\) are \((1,2)\) and \((2,1)\), giving 2 triples. For \(c=2\), the condition becomes \(a+b=5\), with ordered pairs \((2,3)\) and \((3,2)\), giving 2 more triples. Hence, the total number of triples is \(2+2=4\). Remember that reversing an ordered pair produces a different pair.

Correct answer: D. (9)

Explanation: For (a=1,2,3,4,5), the counts of valid (b) are (3,2,2,1,1), totaling (9). Odd sum requires opposite parity.

Correct answer: C. (10)

Explanation: The condition (a+b\le5) remains the same after reversal, so all (10) pairs of (R) qualify. Recognize symmetric conditions.

Correct answer: B. (32)

Explanation: There are (6) pairs in (R), and only ((1,6)) has first component (1). It must be included, so the other (5) pairs are free, giving (2^5=32).

Correct answer: A. (A=B)

Explanation: For non-empty sets, (A\times B=B\times A) only when (A=B). In exams, always check the empty-set condition.

Correct answer: C. ( (4,2) )

Explanation: In ( (4,2) ), the first element (4) is not in (A). In exams, take the first coordinate from (A) and the second from (B).

Correct answer: B. (2)

Explanation: ((A\times B)\cap(C\times D)=(A\cap C)\times(B\cap D)), where (A\cap C={2,3}) and (B\cap D={b}). So there are (2\cdot1=2) elements.

Correct answer: A. (A\times B\subseteq C\times D)

Explanation: Every (a\in A) lies in (C), and every (b\in B) lies in (D), so ((a,b)\in C\times D). In exams, check membership of both coordinates separately.

Correct answer: C. ( (2,4) )

Explanation: (2\in[1,3]) and (4\in(2,5)), so ((2,4)\in A\times B). In exams, check endpoints of closed and open intervals carefully.

Correct answer: A. (6)

Explanation: Pairs with equal coordinates are of the form ((a,a)), and their number is (n(A)). In exams, treat them as diagonal pairs.

Correct answer: C. (42)

Explanation: The direct answer is option C: 42 ordered pairs. If a set has 7 elements, then \(A\times A\) has \(7\times7=49\) ordered pairs, because each first coordinate can be paired with each second coordinate. Among these, 7 pairs have equal coordinates: \((a,a)\), one for each element of A. The required pairs have distinct coordinates, so subtract the equal-coordinate pairs: \(49-7=42\). Option C is correct. Option A, 7, counts only the pairs with equal coordinates, which are specifically excluded. Option B, 14, does not account for all possible choices of two different coordinates. Option D, 49, is the total number of pairs and therefore includes the 7 unwanted diagonal pairs. The order matters in a Cartesian product, so selecting a first and second coordinate creates an ordered pair. Memory cue: distinct-coordinate pairs in \(A\times A\) equal \(n(n-1)\); here \(7\cdot6=42\).

Correct answer: B. (2^{20})

Explanation: A relation is any subset of (A\times B), and (n(A\times B)=20). Hence the number of relations is (2^{20}).

Correct answer: C. (15)

Explanation: The Cartesian product contains every ordered pair whose first member comes from A and whose second member comes from B. Since A has 3 elements and B has 2 elements, the product has 3 × 2 = 6 distinct ordered pairs. A subset is formed by selecting some of these six pairs, and the order in which the selected pairs are listed does not create a new subset.

To obtain a subset with exactly two ordered pairs, choose any 2 pairs from the 6 available pairs. The number of such choices is \\(\binom{6}{2}=\frac{6\times5}{2}=15\\). Therefore, option C is correct. The answer is not 12 or 36: 12 does not represent the required combination count, while 36 is the square of the product size rather than the number of two-element subsets.

Correct answer: B. (9)

Explanation: Since (a^2-b^2=(a+b)(a-b)), the value is (9\cdot1=9). In exams, match coordinates directly from pair equality.

Correct answer: B. (A=\varnothing) or (B=\varnothing)

Explanation: A Cartesian product is empty when at least one set is empty. In exams, both sets need not be empty.

Correct answer: D. (6)

Explanation: The direct answer is option D: there are 6 elements. First form the union. Since B={3} and C={4,5} have no common element, \(B\cup C=\{3,4,5\}\), which has 3 elements. The Cartesian product \(A\times(B\cup C)\) contains every ordered pair whose first entry is from A and second entry is from the union. A has 2 elements, and the union has 3, so the product has \(2\times3=6\) elements. They are \((1,3),(1,4),(1,5),(2,3),(2,4),(2,5)\). Option A, 3, counts only the union and forgets the first factor. Option B, 4, and option C, 5, do not follow the product rule and undercount. Option D, 6, correctly uses \(|X\times Y|=|X||Y|\). In these questions, simplify the union first, then multiply cardinalities.

Correct answer: D. (7)

Explanation: There are (9) total pairs, and the product is odd only for ((1,3)) and ((3,3)). Hence even-product pairs are (9-2=7).