Cartesian product of sets

Since (|A\times B|=12), the number of ways to choose exactly (3) pairs is (\binom{12}{3}=220). Use combinations when an exact size is asked.

A relation is a subset of (A\times B), and (|A\times B|=10), so the number is (2^{10}=1024). Number of relations uses a power of (2).

First find the intersection: \(B\cap C=\{2,5\}\), so \(|B\cap C|=2\) and \(|A|=4\). For a Cartesian product, \(|A\times(B\cap C)|=|A|\,|B\cap C|=4\times2=8\). Therefore, option B is correct. Exam tip: the number of ordered pairs in a Cartesian product equals the product of the cardinalities of the two sets.

((A\times B)\cap(A\times C)=A\times(B\cap C)), and (B\cap C={3}). Hence the cardinality is (2\cdot1=2).

((A\times B)\cup(A\times C)=A\times(B\cup C)), and (B\cup C={4,5,6}). Thus there are (3\cdot3=9) elements.

The correct answer is option C: 6 ordered pairs. The condition |a-b|=1 means that a and b differ by exactly one, whether a is larger or smaller. Check neighboring numbers in {1,2,3,4}. The pairs are (1,2) and (2,1), (2,3) and (3,2), and (3,4) and (4,3). Therefore there are 6 pairs. Option A, 4, usually results from counting only one direction, such as (1,2), (2,3), and (3,4), and perhaps missing one; it is not the full ordered-pair count. Option B, 5, has no complete matching count. Option C is correct because all six listed pairs satisfy the absolute-difference condition. Option D, 8, incorrectly treats every possible pair as neighboring or overcounts. Since Cartesian-product pairs are ordered, (1,2) and (2,1) are different. Memory cue: for n consecutive elements, difference 1 gives 2(n-1) ordered pairs; here 2(3)=6.

The sum \(a+b\) is even when both \(a\) and \(b\) are even or both are odd. Each set contains 2 even numbers, \(0,2\), and 2 odd numbers, \(1,3\). Hence, the number of favourable ordered pairs is \(2\times2+2\times2=8\). Therefore, option B is correct. Exam tip: the sum of two integers is even exactly when they have the same parity.

For (a=1,2,3,4,5), the counts of (b) are (4,4,3,2,1), totaling (14). Include equality in a boundary inequality.

There are (16) total pairs, and (ab) is odd only when both are odd, giving (2\cdot2=4). Hence even products are (16-4=12).

Rewrite the condition as a = 2b + 1. For b = 0, 1 and 2, the corresponding values of a are 1, 3 and 5, respectively. All three values belong to A, giving the ordered pairs (1, 0), (3, 1) and (5, 2). Therefore, the total number of pairs is 3. Exam tip: Substitute every element of one set and verify that the resulting element belongs to the other set.

The difference of two integers is odd exactly when they have opposite parity. Set \(A\) contains 2 even and 2 odd numbers, while set \(B\) contains 2 even and 3 odd numbers. Thus, the number of favourable ordered pairs is \(2\times3+2\times2=10\): either \(a\) is even and \(b\) is odd, or \(a\) is odd and \(b\) is even. Exam tip: remember that \(A\times B\) consists of ordered pairs, so the first component must come from \(A\) and the second from \(B\).

The set \\(A\\) and the set \\(B\\) each contain 5 elements, so their Cartesian product has \\(5\\times5=25\\) ordered pairs. We need the pairs for which \\(a\\ne b\\). It is easier to count the complement first: pairs with \\(a=b\\). Since both sets contain the same five elements, the equal pairs are \\((1,1)\\), \\((2,2)\\), \\((3,3)\\), \\((4,4)\\), and \\((5,5)\\), giving 5 pairs.

Subtract the equal pairs from all pairs: \\(25-5=20\\). Therefore, 20 ordered pairs have unequal components, so option B is correct. The pair order still matters in the Cartesian product, but the equal pairs are exactly the five diagonal pairs. Counting all pairs and removing the unwanted cases is more efficient than listing the 20 required pairs individually.

For every (a\in A), (a^2) is in (B), so (5) pairs are formed. Watch squares of negative numbers carefully.

Checking all sums gives (7) pairs with prime sum. In such questions, look for prime sums like (2,3,5,7).

The condition b/a = 2 can be rewritten as b = 2a, because every element of A is nonzero, so division by a is valid. We then test each possible value of a in A, calculate 2a, and check whether the result belongs to B. A pair is counted only when both components come from their required sets.

For a = 1, b = 2, producing (1,2). For a = 2, b = 4, producing (2,4). For a = 3, b = 6, producing (3,6). Each second component is in B. No other values of a are in A, so there are exactly three valid pairs. Thus option B is correct. The value 8 in B does not produce a further pair because it would require a = 4, which is not in A.

The satisfying pairs are ((3,3),(3,4),(4,3),(4,4)), totaling (4). The correct option should be (C).

For (a=1,2,3,4), the counts of (b) are (5,2,1,1), totaling (9). The correct option should be (B).

The correct answer is option C: C={3,4,5}. In a Cartesian product A×C, each pair has its first component from A and its second component from C. The displayed pairs have first components 1 and 2, matching A={1,2}. Their second components are 3, 4, and 5, and each of these occurs with both first components: (1,3),(1,4),(1,5),(2,3),(2,4),(2,5). Hence C={3,4,5}. Option A, {1,2}, is the first-component set A, not C. Option B, {3,4}, omits 5, which appears in two pairs. Option C contains exactly every second component and is correct. Option D includes all numbers together and does not identify the second-component set. The mention of B={3,4} is not needed for finding C and does not alter the displayed product. Exam cue: in (a,c), read the first positions for A and the second positions for C.

Since (B) is non-empty and (A\times B=\varnothing), we must have (A=\varnothing). An empty product means at least one set is empty.

In ((4,5)), the first component (4) is not in (A). Check the first and second positions separately.

For two finite sets, \(|A\times B|=|A|\cdot|B|\). Here, \(|A|=3\) and \(|A\times B|=15\), so \(3\cdot|B|=15\), giving \(|B|=15\div3=5\). The value 4 results from an incorrect operation or guess. Exam tip: remember that the cardinality of a Cartesian product is the product of the cardinalities of the two sets.

The direct answer is option D, 4095. The Cartesian product has n(A × B) = n(A)n(B) = 4 × 3 = 12 elements. A set with 12 elements has 2^12 subsets, because each element can independently be selected or not selected. This includes the empty subset. The question asks for non-empty subsets, so remove the one empty subset: 2^12 − 1 = 4096 − 1 = 4095. Option A, 12, is only the number of elements in the Cartesian product. Option B, 64, equals 2^6 and uses the wrong exponent. Option C, 255, equals 2^8 − 1, so it corresponds to a different set size. Option D, 4095, correctly counts all subsets except the empty one. A common error is to write 2^12 instead of subtracting 1. The phrase “non-empty” is the signal that the empty subset must be excluded.

The condition gives ((1,3),(2,2),(3,1)), so there are (3) elements. A relation is a selected part of (A\times B).

All elements of (B) are odd, so the sum is odd only when (a) is even. There are (2) even values of (a) and (3) values in (B), totaling (6).

For each \(a\in A\), we must have \(b=10-a\). The possibilities are \(b=9\) for \(a=1\) (not in \(B\)), \(b=8\) for \(a=2\), \(b=7\) for \(a=3\) (not in \(B\)), and \(b=6\) for \(a=4\). Thus, the valid ordered pairs are \((2,8)\) and \((4,6)\), giving a total of \(2\) pairs. Exam tip: because these are ordered pairs, verify that the first component belongs to \(A\) and the second belongs to \(B\).