Cartesian product of sets

The condition \\(|a-b|=3\\) means that the two numbers differ by exactly three, without regard to which one is larger. Since both A and B contain the numbers 1 through 5, we can list the pairs directly. With a smaller first value, the possibilities are (1,4) and (2,5). Reversing each pair gives (4,1) and (5,2).

These four ordered pairs all belong to A × B and satisfy the absolute-difference condition. No other pair works: differences involving adjacent or closer numbers are smaller than 3, and larger gaps are not available within 1 to 5. Because ordered pairs are different when their order is reversed, all four must be counted. Thus option C is correct.

The sum \(a+b\) is even exactly when both \(a\) and \(b\) are even or both are odd. Set \(A\) has 3 even and 2 odd elements, while set \(B\) has 2 even and 2 odd elements. Thus, the number of even-even pairs is \(3\times2=6\), and the number of odd-odd pairs is \(2\times2=4\). Therefore, the total is \(6+4=10\), so option B is correct. The distractor 12 usually results from miscounting one parity group. Exam tip: for an even sum, count the even-even and odd-odd cases separately.

For (a=1,2,3,4,5,6), the counts of (b) are (4,4,4,3,2,1), totaling (18). Include equality in (\le).

The product is odd only when both components are odd. (A) has (3) odd elements and (B) has (2), so (3\cdot2=6).

For each a ∈ A, evaluate b = 3a. For a = 1, 2, 3, and 4, the corresponding values of b are 3, 6, 9, and 12, all of which belong to B. For a = 5, b = 15, which is not in B. Thus the valid ordered pairs are (1, 3), (2, 6), (3, 9), and (4, 12), giving a total of 4 pairs. Exam tip: Count only those values of a for which 3a is an element of B.

From \(a-2b=1\), we get \(a=2b+1\). Substituting \(b=0,1,2,3\) gives \(a=1,3,5,7\), respectively, and all these values belong to \(A\). Thus the valid ordered pairs are \((1,0),(3,1),(5,2),(7,3)\), giving a total of 4 pairs. Exam tip: test every value of one set in the equation and retain only those for which the resulting value belongs to the other set.

The difference of two integers is even exactly when both integers have the same parity—both even or both odd. Set \(A\) has 2 even and 2 odd numbers, while set \(B\) has 2 even and 3 odd numbers. Therefore, the number of valid ordered pairs is \(2\times2+2\times3=4+6=10\). The pairs are ordered, so the first element is selected from \(A\) and the second from \(B\). Exam tip: for an even difference, count pairs with matching parity.

The governing concept is complement counting in a Cartesian product. Since A and B each contain six elements, the total number of ordered pairs in A × B is 6 × 6 = 36. It is easier to count the complement of the required condition. The pairs that do not satisfy a ≠ b are exactly those with a = b: (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6). There are six such diagonal pairs. Therefore the number satisfying a ≠ b is 36 − 6 = 30, so option C is correct. Option D counts every pair and ignores the restriction, whereas options A and B result from subtracting an incorrect number of equal pairs. Ordered-pair order is preserved throughout.

For every (a\in A), (a^2) is in (B), so (7) pairs are formed. Watch squares of negative numbers carefully.

Checking all sums gives (10) pairs with prime sum. In such questions, test possible sums systematically.

The condition b divided by a equals 4 means that b must be four times a. A Cartesian product contains ordered pairs whose first member comes from A and whose second member comes from B. Thus, we must test every value of a in A and calculate its required value of b. This makes the condition easy to check without listing all pairs of the product.

For a=1, b=4, and 4 belongs to B, giving (1,4). For a=2, b=8, giving (2,8). For a=4, b=16, giving (4,16). All three required second components are in B, so there are exactly 3 suitable ordered pairs. Therefore option C is correct. A count of 1 or 2 misses valid pairs, while 4 is impossible because A has only three possible first components.

The satisfying pairs are ((4,4),(4,5),(5,4),(5,5)), so the total is (4). Count each ordered pair once.

The pairs with sum divisible by (5) are ((1,4),(2,3),(3,2),(4,1)). Counting by remainders is easier.

The governing concept is systematic counting of ordered pairs subject to divisibility. Fix each a in A and list its positive multiples that also belong to B. For a = 1, all six members of B qualify: 1, 2, 3, 4, 5, 6, giving six pairs. For a = 2, the qualifying multiples are 2, 4, 6, giving three. For a = 3, they are 3 and 6, giving two. For a = 4, only 4 belongs to B, giving one, and for a = 5, only 5 qualifies, giving one. The total is 6 + 3 + 2 + 1 + 1 = 13. Hence option B is correct. A number such as 8 cannot be used because it is outside B.

In a Cartesian product A × C, every ordered pair has its first component from A and its second component from C. The first components shown are 2 and 4, agreeing with A. The second components appearing in the pairs are 1, 3 and 5, so C = {1,3,5}. This also produces all combinations: each of the two elements of A is paired with each of the three elements of C, giving 2 × 3 = 6 pairs, exactly as listed. Therefore option B is correct. Option A confuses the first component with the second, option C includes elements that never occur as second components, and option D is a set of pairs rather than the required set C.

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

Order inside a set does not matter, so (A=B), and both products are equal. First check whether the two sets are actually the same.

Here, \(|A|=4\). For the Cartesian product, \(|A\times B|=|A|\cdot|B|\). Thus, \(28=4\cdot|B|\), giving \(|B|=28/4=7\). Therefore, the correct answer is 7. Exam tip: the number of ordered pairs in a Cartesian product equals the product of the cardinalities of the two sets.

The direct answer is option C, 4095. The product A × B has n(A × B) = 3 × 4 = 12 elements. For any set with n elements, each element has two choices in a subset—present or absent—so the total number of subsets is 2^n. Here that is 2^12 = 4096. Exactly one of these is the empty subset, so the number of non-empty subsets is 4096 − 1 = 4095. Option A, 255, equals 2^8 − 1 and uses the wrong product size. Option B, 1023, equals 2^10 − 1 and is also based on an incorrect exponent. Option C, 4095, uses 12 correctly and removes the empty subset. Option D, 8191, equals 2^13 − 1, so it is one exponent too large. The key sequence is multiply first, use 2^n, then subtract 1 for “non-empty.”

The condition gives ((1,4),(2,3),(3,2),(4,1)), so there are (4) elements. A relation is a subset of (A\times B).

The sum of an even number and an odd number is always odd. In parity questions, listing all pairs is not necessary.

For (a=0,1,2,3), the counts of (b) are (5,4,3,2), totaling (14). In a strict inequality, the boundary is not included.

From \(a+b=11\), we get \(b=11-a\). Checking the elements of \(A\), the valid ordered pairs are \((1,10),(3,8),(5,6)\), and each second component belongs to \(B\). Thus, there are 3 pairs. Exam tip: Since 11 is odd and every element of \(B\) is even, \(a\) must be odd; therefore, only \(1,3,5\) need to be checked.

Direct answer: Option C, 10. A relation here contains ordered pairs satisfying \\(a\\le b\\), where both numbers are from 1, 2, 3, 4. For a=1, all four b-values work: 4 pairs. For a=2, b=2,3,4 work: 3 pairs. For a=3, b=3,4 work: 2 pairs. For a=4, only b=4 works: 1 pair. The total is \\(4+3+2+1=10\\). Option A, 6, misses several pairs; B, 8, also undercounts; C is correct. D, 12, overcounts pairs that do not satisfy the condition. The equality sign is important: \\(a=b\\) is allowed because the symbol is \\(\\le\\), not just <. Memory cue: list each row from the diagonal to the right.

Independent calculation shows a problem: the supplied answer C=9 is not correct. Count multiples in B for each a. For a=2, the values 4, 6, 10, and 12 are divisible by 2, giving 4 pairs. For a=3, 6, 12, and 15 work, giving 3 pairs. For a=5, 10 and 15 work, giving 2 pairs. Therefore the total is \\(4+3+2=9\\), actually matching Option C. Option A says 7 and undercounts; B says 8 and also undercounts; C is correct because all valid pairs are included; D says 10 and overcounts. The supplied short explanation is accurate: count multiples separately for each a. The wording is clear, so no correction is needed.