Cartesian product of sets

For (x=2) there is (1) choice, for (x=4) there are (3), and for (x=6) there are (4). Hence the total is (1+3+4=8), so choose the option (8).

For each \(x\in A\), the equation \(y=x^2\) gives the pairs \((0,0),(1,1),(2,4),(3,9)\). In every pair, the first component belongs to \(A\) and the second component belongs to \(B\), so all four pairs are in \(A\times B\). Therefore, the answer is \(4\). The value 3 misses one valid pair, while 5 counts an extra pair. Exam tip: square each element of \(A\) and count only those results that belong to \(B\).

The sum is even when both numbers have the same parity. Odd-odd gives (4) pairs and even-even gives (1), so total is (5).

(x+y=5), and both coordinates must lie from (1) to (4). Thus the four pairs are ((1,4),(2,3),(3,2),(4,1)).

In a Cartesian product B×A, order is essential. The first coordinate must be an element of B, and the second coordinate must be an element of A. Here B contains 1 and 2, while A contains a, b, and c. Therefore examples of members include (1,a), (1,b), (1,c), (2,a), (2,b), and (2,c).

The pair (1,a) follows the required order: 1 is taken from B and a is taken from A. Hence option B is correct. The pairs (a,1), (b,2), and (c,1) have the elements in the reverse order, because their first coordinates come from A and their second coordinates come from B. They could belong to A×B, but not to B×A. This illustrates why Cartesian products are ordered products.

We need ordered pairs (x,y) in A×B such that y is the square of x and x is odd. First identify the odd elements of A. They are 1, 3, and 5. For each of these values, calculate its square and check whether the result belongs to B. This simultaneously checks both the equation and the Cartesian-product requirement.

For x=1, y=1, and 1 is in B, so (1,1) works. For x=3, y=9, and 9 is in B, so (3,9) works. For x=5, y=25, and 25 is in B, so (5,25) works. Thus three pairs satisfy all conditions. The even values 2 and 4 are excluded because x must be odd. Therefore option B, 3, is correct.

The prime sums possible are (3,5,7). The valid pairs are ((1,2),(1,4),(2,3),(3,2),(3,4)), so there are (5) pairs.

For (x=1) there are (4), for (x=2) there are (4), for (x=3) there is (1), and for (x=4) there are (2). Total is (11).

For (x=2), (4,6,10) work; for (x=3), (6,15) work; for (x=5), (10,15) work. The total is (3+2+2=7), so the correct option is (7).

The direct answer is option B: 5 pairs. We need ordered pairs \((x,y)\) with \(x,y\in\{1,2,3,4\}\) and \(x+y\) divisible by 3. Possible sums are 3, 6, or 9, but with values from 1 to 4, only 3 and 6 occur. Sum 3 gives \((1,2),(2,1)\). Sum 6 gives \((2,4),(3,3),(4,2)\). Thus the total is \(2+3=5\). Option A, 4, misses one valid pair. Option B, 5, lists all valid ordered pairs and is correct. Option C, 6, and option D, 7, include extra pairs whose sums are not divisible by 3. Order matters: \((1,2)\) and \((2,1)\) are different pairs. A useful exam cue is to group numbers by their remainders modulo 3, but direct checking here is quickest.

The equation (xy-x-y=0) gives ((x-1)(y-1)=1). Among positive integers, only ((2,2)) works, so the count is (1).

(n(A\times B)=12), and ((A\times B)\cap(C\times D)=(A\cap C)\times(B\cap D)) has (4) elements. So the difference has (12-4=8) elements.

The first coordinates are (1,4) and the second coordinates are (2,3). Order is very important in (A\times B).

(n(A\times B)=3\cdot2=6) and then (6\cdot4=24). An ordered pair is treated as one element in the next Cartesian product.

(n(A\times B)=mn) and (n(B\times A)=nm), so they are equal. Equal cardinality does not necessarily mean equal sets.

If (A=B), both products contain exactly the same ordered pairs. Equal number of elements alone is not enough.

The only common ordered pair in both Cartesian products is ((2,2)). In ordered pairs, ((1,2)) and ((2,1)) are different.

(B={2,3,5,7}); the sum is even when both have the same parity. Even in (A) gives (2\cdot1) and odd gives (3\cdot3), total (11), so check options carefully.

For (a=1) there are (3), for (a=2) there are (3), for (a=3) there is (1), and for (a=4) there is (1). Total is (8), so test each first coordinate separately.

For (ab=0), (a=0) gives (3) pairs and (b=0) gives (3) pairs, but ((0,0)) is counted twice. Hence total is (3+3-1=5).

For (a=1) there are (3), for (a=2) there are (2), and for (a=3) there is (1) choice. Total is (6), and the order in inequality matters.

For (a=2), only (9); for (a=3), (4,10); for (a=5), (4,6,9), giving total (6). Therefore the correct count should be (6), so the listed options do not match.

(n(A\times B)=12), and the common part ((A\cap B)\times(A\cap B)) has (4) pairs. So the difference has (12-4=8) pairs.

(B={-2,2}), and the pairs are ((2,-2)) and ((-2,2)). Keep the sets of both coordinates separate while applying the equation.

Cartesian product distributes over union in the second component. Remember (A\times(B\cup C)=(A\times B)\cup(A\times C)).