Reflexive relation

Step 1: A reflexive relation must contain ((1,1),(2,2),(3,3),(4,4)). Step 2: In (P), ((2,2)) is already counted among these four, while ((1,3)) and ((3,1)) are two extra fixed pairs. Step 3: So 6 pairs are fixed, (16-6=10) pairs are free, and the count is (2^{10}).

Step 1: On 6 elements, 6 self-pairs are compulsory. Step 2: To have 9 total pairs, choose 3 pairs from the (36-6=30) non-self pairs. Step 3: Hence the number is (\binom{30}{3}).

Step 1: On 4 elements, the total number of reflexive relations is (2^{16-4}=2^{12}). Step 2: The identity relation is exactly one of them. Step 3: Only the identity relation is removed, so the count is (2^{12}-1).

Step 1: Four self-pairs are fixed by reflexivity. Step 2: At most 6 total pairs means choosing 0, 1, or 2 pairs from the 12 non-self pairs. Step 3: The count is (\binom{12}{0}+\binom{12}{1}+\binom{12}{2}=1+12+66=79).

Step 1: The 3 self-pairs are compulsory. Step 2: Choose exactly 1 of the given 3 non-self pairs in (\binom{3}{1}=3) ways. Step 3: The remaining 3 non-self pairs are free, so the count is (3\times2^3=24).

Step 1: Since (R) is reflexive, ((a,a)\in R) for every (a\in A). Step 2: In (A\times A-R), all pairs of (R) are removed, so the self-pairs are removed. Step 3: On a non-empty set, a relation without self-pairs cannot be reflexive.

Step 1: Since (R) is reflexive, all ((a,a)) are in it. Step 2: The inverse of ((a,a)) is also ((a,a)), so these pairs are in (R^{-1}) too. Step 3: The intersection keeps all self-pairs and (R^{-1}\cap R) is also symmetric.

Step 1: Reflexivity of (R) gives (aRa). Step 2: Reflexivity of (S) gives (aSa). Step 3: Taking the middle element as (a), we get (a(S\circ R)a), so the composition is reflexive.

Step 1: The 4 self-pairs are fixed. Step 2: For 12 total pairs, 8 non-self pairs are needed, meaning 4 unordered pairs are chosen in both directions. Step 3: Choose 4 of the 6 unordered pairs in (\binom{6}{4}=15) ways.

Step 1: The equal-remainder groups modulo 4 are ({1,5}), ({2,6}), ({3,7}), and ({4,8}). Step 2: Each group gives (2^2=4) ordered pairs. Step 3: Total pairs are (4+4+4+4=16), including all self-pairs.

Step 1: Square remainders are grouped as ({0}) with remainder 0, ({1,5}) with remainder 1, ({2,4}) with remainder 4, and ({3}) with remainder 3. Step 2: Ordered pairs are formed within equal square-remainder groups. Step 3: The total is (1^2+2^2+2^2+1^2=10).

Step 1: The power set has (2^4=16) elements. Step 2: A reflexive pair is ((X,X)) for each element (X) of the power set. Step 3: Hence the number of reflexive pairs is 16.

Step 1: For (X\subseteq Y), each basic element has three choices: in (Y) only, in both, or in neither. Step 2: Being in (X) only is not allowed. Step 3: For 4 basic elements, the total number of pairs is (3^4=81).

Step 1: For reflexivity, every ((a,a)) must satisfy the rule. Step 2: The self-sums are (4,8,12,16). Step 3: The largest self-sum is 16, so the minimum (k) is 16.

Step 1: For self-pairs, check (a^2+a^2=2a^2). Step 2: The largest value occurs at (a=\pm2), giving (2\cdot4=8). Step 3: The inequality is strict, so (8<k) is needed, and the minimum integer value is (k=9).

Step 1: Count the divisors present in (A) for each second component. Step 2: For (1,2,4,8,16), the counts are (1,2,3,4,5). Step 3: The total number of pairs is (1+2+3+4+5=15).

Step 1: The equal absolute value groups are ({-4,4}), ({-2,2}), and ({0}). Step 2: Total pairs are (2^2+2^2+1^2=9). Step 3: Since 5 are self-pairs, the number of non-self pairs is (9-5=4).

Step 1: Since (R) is reflexive, (I_A\subseteq R). Step 2: The union of a set with its subset is the larger set itself. Step 3: Therefore (R\cup I_A=R).

Step 1: A reflexive (R) contains the whole identity relation (I_A). Step 2: Intersecting with (I_A) keeps exactly those self-pairs. Step 3: Therefore (R\cap I_A=I_A).

Step 1: In composition, ((1,2)) and ((2,3)) can be joined because the middle element 2 matches. Step 2: They produce ((1,3)) in (R\circ R). Step 3: In composition questions, first look for consecutive connectable pairs.

Step 1: The 4 self-pairs are fixed. Step 2: To have 7 total pairs, choose (7-4=3) non-self pairs. Step 3: There are 12 non-self pairs, so the count is (\binom{12}{3}).

Step 1: A reflexive relation has (n) compulsory self-pairs. Step 2: To have exactly (n+3) pairs, 3 additional non-self pairs are needed. Step 3: Choose 3 from (n^2-n) non-self pairs, so the count is (\binom{n^2-n}{3}).

Step 1: The 3 self-pairs are fixed. Step 2: For ((1,2),(2,1)), valid choices are neither, only ((1,2)), or only ((2,1)); both together are not allowed. Step 3: The remaining 4 non-self pairs are free, so the count is (3\times2^4=48).

Step 1: (R) contains all self-pairs, so they remain in the union. Step 2: In (R\cup R^{-1}), every pair appears with its reverse. Step 3: Therefore the relation is both reflexive and symmetric.

Step 1: Since (R) is reflexive, all self-pairs are in (R). Step 2: Because (R\subseteq S), all those self-pairs are also in (S). Step 3: Therefore (S) is definitely reflexive.