Cartesian product of sets

For (a=1,2,3,4,5), the counts of (b) are (6,5,3,2,2), totaling (18). In product conditions, the boundary changes.

Direct answer: Option C, 18. A has 2 elements, B has 3, and C has 3. First, \\(|A\\times B|=2\\cdot3=6\\). Next, each of these 6 ordered pairs can be combined with any of the 3 elements of C, so \\(|(A\\times B)\\times C|=6\\cdot3=18\\). Equivalently, multiply all three cardinalities: \\(2\\cdot3\\cdot3=18\\). Option A, 12, does not include all three choices correctly. B, 15, is also not the product of the given sizes. C is correct. D, 24, is too large and would require a different set size. The parentheses change the structure of the objects, but not this counting multiplication. Memory cue: count choices at each position and multiply them.

There are (25) total pairs and (5) diagonal pairs, so (25-5=20). Removing the diagonal leaves pairs with (a\ne b).

The sum of two integers is divisible by 2 exactly when both integers have the same parity—both even or both odd. There are 3 even elements in \(A\) and 2 in \(B\), giving \(3\times2=6\) pairs. Similarly, there are 3 odd elements in \(A\) and 2 in \(B\), giving another \(3\times2=6\) pairs. Thus, the total is \(6+6=12\), so option B is correct. Mixed-parity pairs have an odd sum and must not be counted. Exam tip: For parity-based counting, first separate the elements of each set into even and odd groups.

For (a=1,2,3,4), the counts of (b) are (3,2,1,0), totaling (6). Apply the condition to each first component.

The direct answer is 19, so option B is correct. A relation T is a collection of ordered pairs from the Cartesian product A×B. Since A and B each have five elements, A×B has 5×5=25 ordered pairs. We need to retain pairs for which the absolute difference between the two coordinates is at most 2. It is quicker to count the pairs that fail the condition, namely those with |a−b|>2. These are (1,4), (1,5), (2,5), (4,1), (5,1), and (5,2), giving six excluded pairs. Therefore |T|=25−6=19. Option A, 17, excludes too many pairs. Option C, 21, excludes only four pairs and misses two invalid pairs. Option D, 23, excludes only two pairs. Remember that ≤ includes equality, so differences 0, 1, and 2 are all allowed; for example, (1,3) belongs to T.

The governing algebraic step is to transform the condition before counting. Starting with a+b=ab, rearrange to ab−a−b=0. Adding 1 to both sides in factored form gives (a−1)(b−1)=1. Since a and b belong to {1,2,3,4,5,6}, both factors are nonnegative integers. The only way two such integers can have product 1 is that each equals 1. Hence a−1=1 and b−1=1, giving a=2 and b=2. The single valid ordered pair is (2,2), so the answer is 1 and option B is correct. Options C and D incorrectly count extra pairs that do not satisfy the original equation.

The governing concept is the union rule for sets of ordered pairs. First list the pairs satisfying a+b=9: (3,6), (4,5), (5,4), and (6,3), giving four pairs. Next list the pairs satisfying ab=12: (2,6), (3,4), (4,3), and (6,2), also giving four pairs. The two lists have no common pair. Indeed, the products of the sum-9 pairs are 18 or 20, not 12, while the factor pairs of 12 do not have sum 9. Hence the union contains 4+4−0=8 ordered pairs. Option B is correct. The other choices either omit a valid pair or add an overlap that does not exist.

The governing concept is the structure of a nested Cartesian product, where parentheses are significant. An element of A×B has the form (a,b), so (1,3) is an element of A×B. An element of (A×B)×C must then be an ordered pair whose first coordinate is an entire element of A×B and whose second coordinate is an element of C. Its form is therefore ((a,b),c). Since (1,3)∈A×B and 5∈C, ((1,3),5) belongs to the required product, so option B is correct. Option A flattens the nested pair into a triple, option C groups the factors as A×(B×C), and option D reverses the factor order. These structures are not interchangeable merely because they contain the same numbers.