Cartesian product of sets

In (B\times A), the first component is from (B) and the second from (A). Do not confuse (A\times B) with (B\times A).

In ordered pairs, ((1,3)) and ((3,1)) are not the same. Hence (A\times B) is generally not equal to (B\times A).

(|A\times B|=2\cdot1=2), then (|(A\times B)\times C|=2\cdot3=6). Cardinalities multiply in nested products too.

Check all three values of \(x\): \((-1)^2=1\) gives \((-1,1)\), \(0^2=0\) gives \((0,0)\), and \(1^2=1\) gives \((1,1)\). Thus, there are 3 ordered pairs. Option 6 is the total number of pairs in \(A\times B\), but not all of them satisfy \(x^2=y\). Exam tip: For this type of question, compute \(x^2\) for each \(x\) and verify whether the resulting \(y\) belongs to \(B\).

For (a=2,3,4), the counts are (1,2,3), giving (6). For inequalities, count using the first component.

For the Cartesian product, |A×B|=|A|×|B|. Here, |A|=3 and |A×B|=12, so 3×|B|=12 and hence |B|=12÷3=4. The value 12 is the cardinality of the product, not of B. In an exam, first find |A| and divide the product cardinality by it.

(|A\times B|=mn), so the number of subsets is (2^{mn}). First find the cardinality of the base set.

A relation is any subset of (A\times B), and (|A\times B|=6). Hence the number of relations is (2^6=64).

(|A\times B|=3\cdot2=6), so non-empty subsets are (2^6-1=63). Do not forget to subtract the empty set.

Direct answer: Option B, \\(B=\\{a,b\\}\\). A Cartesian product contains ordered pairs \\( (u,v) \\) where the first component comes from the first set and the second component comes from the second set. Since the first components are 1 and 2, they match A. The second components appearing are a and b, so B must be \\(\\{a,b\\}\\). Option A repeats A, not B. Option B correctly collects the second entries. Option C lists pairs, so it is a subset of the product rather than the set B. Option D mixes first and second components into one set and does not describe the second factor. The order matters: \\(A\\times B\\) is not generally the same as \\(B\\times A\\). Memory cue: in every ordered pair, first coordinate belongs to A and second coordinate belongs to B.

If (A) is non-empty and (A\times B=\varnothing), then (B=\varnothing). A Cartesian product is empty if at least one set is empty.

The first set is empty, so no ordered pair can be formed. Treat the empty set as a set with (0) elements.

Since (R) can be at most the whole (A\times B), and (|A\times B|=20). If maximum elements are asked, do not write (2^{20}).

For Cartesian product of three sets, the cardinality is (2\cdot3\cdot4=24). For ordered triples, multiply all three sizes.

(B\cup C={3,4,5}), so (|A\times(B\cup C)|=2\cdot3=6). First evaluate the set inside the brackets.

((A\times B)\cap(A\times C)=A\times(B\cap C)), and (B\cap C={3}). Thus the pairs are ((1,3)) and ((2,3)).

(|A\times B|=6), and choosing (2) elements gives (\binom{6}{2}=15). Exactly (2) elements means a combination.

(|A\times B|=6), so there are (6) ways to choose a one-element relation. A singleton relation is just one ordered pair.

The sum of two numbers is even when they have the same parity—both must be even or both must be odd. Each set contains 2 even elements, \(\{2,4\}\), and 2 odd elements, \(\{1,3\}\). Hence, the number of favourable ordered pairs is \(2\times2+2\times2=8\). Exam tip: remember that an even sum requires equal parity, while an odd sum requires opposite parity. The value 16 would count every pair in \(A\times B\), including pairs with one even and one odd component.

The pairs are ((1,5),(2,4),(3,3),(4,2),(5,1)), so there are (5). Reversed pairs count separately in ordered pairs.

For (a=1,2,3,4), the counts are (4,2,1,1), totaling (8). In divisibility, check each (a) separately.

The direct answer is option C, 3 pairs. A pair (a,b) belongs to A × B only when a comes from A and b comes from B. We must test the rule b = a^2 for each a in A. For a = 1, b = 1^2 = 1, and 1 is in B, giving (1,1). For a = 2, b = 2^2 = 4, and 4 is in B, giving (2,4). For a = 3, b = 3^2 = 9, and 9 is in B, giving (3,9). Thus there are three valid pairs. Option A, 1, misses two valid cases. Option B, 2, also stops before checking all three values of a. Option C, 3, correctly counts all successful checks. Option D, 6, is the total number of pairs in A × B, but not every pair satisfies b = a^2. A useful exam method is to calculate the required b for every a and then check whether that b belongs to B.

The direct answer is option B: 6 pairs. We count by fixing a and then listing b values that make a + b < 3. If a = 0, then b < 3, so b = 0, 1, 2: 3 choices. If a = 1, then b < 2, so b = 0, 1: 2 choices. If a = 2, then b < 1, so b = 0: 1 choice. The total is 3 + 2 + 1 = 6. The valid pairs are (0,0), (0,1), (0,2), (1,0), (1,1), and (2,0). Option A, 5, misses one pair. Option B, 6, matches the complete count. Option C, 7, usually results from including a boundary value such as a + b = 3, but the condition is strictly less than 3, not less than or equal to 3. Option D, 8, overcounts still more pairs. Memory cue: for “less than,” do not include equality; count each row carefully.

The diagonal pairs are ((1,1),(2,2),(3,3)), so the count is (3). For equal sets, the diagonal count is usually (|A|).

The direct answer is option C, 3 pairs. The condition a/b = 2 can be rewritten by multiplying both sides by b: a = 2b. Now test each b in B. For b = 1, a = 2, which is in A, so (2,1) works. For b = 2, a = 4, which is in A, so (4,2) works. For b = 3, a = 6, which is in A, so (6,3) works. Therefore there are three valid ordered pairs. Option A, 1, counts only one pair and ignores the other two. Option B, 2, also misses one valid pair. Option C, 3, includes every allowed value of b and is correct. Option D, 4, has no fourth possible value because B contains only 1, 2, and 3; alternatively, checking a = 2, 4, 6 gives exactly the same three pairs. The important idea is that ordered pairs must satisfy both membership and the equation. Converting a fraction condition into a = 2b makes the counting easy.