Cartesian product of sets

There are (12) pairs, and after forcing (2) in and (1) out, (9) pairs remain free. Hence the number is (2^9=512).

(|A\times B|=10), so choosing exactly (5) pairs gives (\binom{10}{5}=252). Use combinations for exact-size subsets.

The condition gives ((8,3)) and ((5,4)), so (|R|=2). Check possible (b)-values systematically.

The pairs are ((0,5),(3,4),(4,3),(5,0)), so the count is (4). Reversed ordered pairs are counted separately.

For (a=1), there are (5) pairs; for (a=2), (2); and for (a=3), (1), totaling (8). Do not include (a=b).

Apply both conditions simultaneously. For \(a=1\), the values \(b=2,4,6\) give the prime sums \(3,5,7\), contributing 3 pairs. For \(a=2\), \(b=3,5\) contribute 2 pairs; for \(a=3\), \(b=4\) contributes 1 pair; for \(a=4\), \(b=7\) contributes 1 pair; and for \(a=5\), \(b=6\) contributes 1 pair. Thus the total is \(3+2+1+1+1=8\) pairs. Therefore, option B is correct. Exam tip: for each \(a\), list only elements \(b>a\), then test whether \(a+b\) is prime.

The complement contains pairs with odd sum, and the count is (2\cdot2+2\cdot2=8). For complements, write the opposite condition.

The direct answer is option B: \(|R|=12\). For each \(a\in A\), count the values \(b\in B\) satisfying \(b\ge a^2-1\). If \(a=0\), the lower bound is \(-1\); every element 0,1,2,3,4 of B works, so there are 5. If \(a=1\), the lower bound is 0; again all 5 values work. If \(a=2\), the lower bound is 3; only 3 and 4 work, giving 2. If \(a=3\), the lower bound is 8; no element of B reaches 8, giving 0. Thus \(|R|=5+5+2+0=12\). Option A, 11, undercounts by one. Option B, 12, matches the complete count. Option C, 13, adds an invalid pair, probably by mishandling the boundary. Option D, 14, is also too large because B has only five elements and the last case contributes none. Always calculate the bound separately for each \(a\).

The components must have the same parity and also (\gcd(a,b)=1), giving (7) pairs. Check (\gcd) carefully among same-parity pairs.

First, \(B\setminus C=\{1,4\}\), so \(\lvert A\times(B\setminus C)\rvert=3\times2=6\). Also, \((A\times B)\cap(A\times C)=A\times(B\cap C)\), and \(B\cap C=\{2,3\}\), so this intersection also contains \(3\times2=6\) ordered pairs. Therefore, the total is \(6+6=12\). Exam tip: simplify \((A\times B)\cap(A\times C)\) as \(A\times(B\cap C)\) before counting elements.

Counting opposite-parity pairs under the product bound gives (10). For combined conditions, check the bound first and then parity.

Checking possible differences gives (6) pairs with prime difference. Remember that (1) is not prime.

For every pair \((b,c)\), the equation \(a=b+c\) determines exactly one value of \(a\). When \(c=0\), the values of \(a\) are \(1,2,3\); when \(c=1\), they are \(2,3,4\). All these values belong to \(A\), so the total number of ordered triples is \(3+3=6\). Repeated values of \(a\) still correspond to different triples because their \((b,c)\) pairs are different. In an exam, count the possible \((b,c)\) pairs first and then check whether the resulting \(a\) lies in \(A\).

For (\operatorname{lcm}(a,b)=6), both values must divide (6), and there are (9) pairs. Check using the list of divisors.

The ordered pairs satisfying a<b are (1,2), (1,3), (1,4), (2,3), (2,4), and (3,4), giving 6 pairs. The pairs satisfying a+b=5 are (1,4), (2,3), (3,2), and (4,1), giving 4 pairs. The pairs (1,4) and (2,3) satisfy both conditions, so they must not be counted twice. Therefore, the required number is 6+4−2=8. Exam tip: For two conditions joined by ‘or’, use |P∪Q|=|P|+|Q|−|P∩Q|.

There are (15) pairs with (a+b\le4), and equal pairs among them are ((0,0),(1,1),(2,2)). Hence (15-3=12).

There are (12) pairs with sum divisible by (3), and equal pairs among them are ((3,3),(6,6)). Hence (12-2=10).

For \(a=1\), there are 6 possible multiples of \(a\) in \(B\): 1, 2, 3, 4, 5, 6. For \(a=2\), there are 3: 2, 4, 6; and for \(a=3\), there are 2: 3, 6. Thus, the total number of pairs before applying \(b\ne a\) is \(6+3+2=11\). The three pairs \((1,1),(2,2),(3,3)\) have \(b=a\), so they must be excluded. Therefore, the required number is \(11-3=8\). Exam tip: count the available multiples for each value of \(a\), then remove the equal-coordinate pairs.

The direct answer is option A, 4. For each a in A, calculate a² and find the members b of B satisfying b ≡ a² (mod 5). Since B is {0,1,2,3,4,5}, note that both 0 and 5 are congruent to 0 modulo 5, while each of 1, 2, 3, and 4 has one representative. For a = 1, a² = 1, so b = 1: one pair. For a = 2, a² = 4, so b = 4: one pair. For a = 3, a² = 9 ≡ 4, so b = 4: one pair. For a = 4, a² = 16 ≡ 1, so b = 1: one pair. Thus the pairs are (1,1), (2,4), (3,4), and (4,1), giving 4. Option B, 5, adds an unsupported pair. Option C, 6, and option D, 8, overcount. Do not count different a-values as one pair merely because they produce the same b.

There are (8) pairs with odd sum, so subsets of (R) are (2^8=256). First find the cardinality of the relation.

Even sum requires same parity, along with (a<b). Odd values give (3) pairs and even values give (1), totaling (4).

There are (3) primes and (2) composites, so the count is (3^2+2^2=13). The number (1) is neither prime nor composite.

The relation contains ordered pairs from A and B for which the difference a minus b is either 1 or 3. We can count the pairs for each difference separately. Since a pair cannot have both differences at once, the two counts can be added. For a fixed a, the required b is determined, and it must belong to B.

For a-b=1, the valid pairs are (1,0), (2,1), (3,2), (4,3), and (5,4), giving 5 pairs. For a-b=3, the valid pairs are (3,0), (4,1), and (5,2), giving 3 pairs. No other values work within the given sets. Hence |R|=5+3=8, so option C is correct. The count must not include values of b outside B.

The diagonal pairs are ((1,1),(2,2),(3,3),(4,4)), so (4) must be removed. The number of diagonal pairs is (|A|).

There are (7) ordered triples from (1) to (3) with sum (6). In ordered triples, order is counted separately.