Cartesian product of sets

For each element of \(A\), calculate \(b=2a\): when \(a=1\), \(b=2\); when \(a=2\), \(b=4\); and when \(a=3\), \(b=6\). All three resulting values belong to \(B\), giving the ordered pairs \((1,2),(2,4),(3,6)\). Hence, the total is \(3\). Choosing \(4\) is incorrect because \(A\) has only three possible values of \(a\). Exam tip: list the ordered pairs satisfying the condition and then count them.

For (b=1,2,3), the possible values of (a) are (1,2,3), totaling (6). Sometimes counting by the second component is easier.

The correct answer is option B: 3 pairs. A Cartesian product contains ordered pairs, with the first entry from A and the second from B. Since both sets are {1,2,3}, list the pairs whose sum is at least 5. For a=1, the largest sum is 1+3=4, so none work. For a=2, only (2,3) gives 5. For a=3, (3,2) gives 5 and (3,3) gives 6. Thus the valid pairs are (2,3), (3,2), and (3,3), giving 3. Option A, 2, misses one valid pair. Option B, 3, is correct. Option C, 4, counts an extra pair such as (1,3), whose sum is only 4. Option D, 5, also overcounts. The symbol >= includes equality, so sums exactly 5 must be included. Memory cue: check every first entry and include the boundary value.

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

Here (A=B), so (A\times B) and (B\times A) are the same. Equality is clear when the two sets are identical.

((1,4)\in A\times B), but ((1,4)\notin B\times A) because (1\notin B). One counterexample is enough.

Since \(A\) and \(B\) have no common element, \(A\cup B=\{1,2,3,4\}\) and \(|A\cup B|=4\). Also, \(|C|=2\). For Cartesian products, \(|X\times Y|=|X|\cdot|Y|\); therefore, \(|(A\cup B)\times C|=4\times2=8\). Option 16 incorrectly multiplies the cardinalities of all three original sets. Exam tip: find the size of the union first, then apply the Cartesian-product rule.

((A\times B)\cup(A\times C)=A\times(B\cup C)), and (B\cup C={2,3,4}). Hence the cardinality is (2\cdot3=6).

The condition is \\(a-b=1\\), which can be rewritten as \\(a=b+1\\). We test the two possible values of \\(b\\) from \\(B=\\{1,2\\}\\). When \\(b=1\\), the required value is \\(a=2\\), and 2 belongs to \\(A=\\{1,2,3\\}\\). When \\(b=2\\), the required value is \\(a=3\\), which also belongs to \\(A\\).

The resulting ordered pairs are \\((2,1)\\) and \\((3,2)\\). Both belong to \\(A\\times B\\) and satisfy the difference condition. There are no other choices for \\(b\\), so the count is 2. Option B is therefore correct. It is important not to count \\((1,2)\\), because its difference is \\(1-2=-1\\), not 1; reversing an ordered pair changes the condition.

For each \(a\in A\), calculate \(b=2^a\): when \(a=1,2,3,4\), the corresponding values are \(2,4,8,16\). The first three values belong to \(B\), giving the ordered pairs \((1,2),(2,4),(3,8)\). Since \(16\notin B\), \((4,16)\) is not in \(A\times B\). Therefore, the total number of valid pairs is 3. Exam tip: For this type of question, evaluate the given rule for every first component and retain only those pairs whose second component belongs to the second set.

Checking systematically gives (6) pairs with prime sum. In such questions, test all possible sums in an organized way.

The sum of two numbers is odd only when one number is even and the other is odd. Set \(A\) has 2 even and 2 odd elements, while set \(B\) has 2 even and 1 odd element. Hence, the number of suitable ordered pairs is \((2\times1)+(2\times2)=2+4=6\). As an exam tip, count separately the cases where the first component is even and where it is odd, since the pairs are ordered.

For (a=1,2,3), the possible values of (b) are (4,3,3), totaling (10). In product conditions, the limit changes for each (a).

The direct answer is option C, 11. Two numbers are coprime when their greatest common divisor is 1. Count systematically by fixing the first number a. If a = 1, all four values of b work: 1,2,3,4, giving 4 pairs. If a = 2, b = 1 and 3 work, giving 2 pairs; b = 2 and 4 share a factor 2. If a = 3, b = 1,2,4 work, giving 3 pairs; b = 3 shares factor 3. If a = 4, b = 1 and 3 work, giving 2 pairs; b = 2 and 4 are not coprime. Total = 4 + 2 + 3 + 2 = 11. Option A, 9, misses two valid pairs. Option B, 10, also undercounts. Option C, 11, matches the complete count. Option D, 12, overcounts two non-coprime pairs. Remember that 1 is coprime with every positive integer, and coprime does not mean the numbers themselves must be prime.

The difference of two integers is even exactly when they have the same parity—both even or both odd. Set \(A\) has 1 even and 2 odd elements, while set \(B\) has 2 even and 2 odd elements. Hence, the number of favourable ordered pairs is \(1\times2+2\times2=6\). Exam tip: For an even difference, count pairs formed by elements of matching parity.

For (a=1,2,3,4), the possible values of (b) are (4,3,2,1), totaling (10). Make rows according to the boundary in such counts.

The condition (|a-b|=2) gives the pairs ((1,3),(2,4),(3,1),(4,2)). In ordered pairs, reversed pairs are counted separately.

In (A\times B), the first component comes from (A) and the second from (B). Changing order changes the ordered pair.

For product of three sets, the cardinality is (3\cdot2\cdot3=18). For ordered triples, multiply the sizes of all sets.

For each element of A, calculate b = 8 − a. When a = 1, b = 7, which is not in B; when a = 2, b = 6, which is in B; when a = 3, b = 5, which is not in B; and when a = 4, b = 4, which is in B. Thus, the valid pairs are (2, 6) and (4, 4), giving a total of 2 pairs. Exam tip: In A × B, the first element must come from A and the second from B.

For (a=0,1,2,3), the counts of (b) are (4,3,2,1), totaling (10). Count row-wise in inequality questions.

For (b=1), (a=2,3,4,5); for (b=2,3), counts are (2,0), so total is (6). The correct option should be (D).

For every (a\in A), (a^2\in B), so (4) pairs are formed. In rule-based questions, check all first components.

For (a=1), there are (4) pairs; for (a=2), (2); and for (a=3), (1), totaling (7). Check each divisor separately.

The pairs are ((2,2),(2,4),(4,2)), so the number is (3). In (\gcd) questions, check common factors.