Match the LIST-I with LIST-II.
Each of the following relations is defined on set {1, 2, 3, 4}. Match the relation types.

LIST-I (Relation)	LIST-II (Relation type)
A. {(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)}	I. Antisymmetric
B. {(2,4), (4,2)}	II. Only transitive
C. {(1,2), (2,3), (3,4)}	III. Only symmetric
D. {(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)}	IV. neither symmetric, nor transitive

Choose the correct answer from the options given below:

Explanation

Relation A = {(2,2),(2,3),(2,4),(3,2),(3,3),(3,4)} contains both (2,3) and (3,2) with 2≠3, so it is not antisymmetric, and it contains (2,4) without (4,2), so it is not symmetric; but checking every composition (e.g. (2,3)&(3,2)→(2,2), (2,3)&(3,4)→(2,4), (3,2)&(2,4)→(3,4)) shows the required pair is always present, so A is transitive — 'only transitive' (II). Relation B = {(2,4),(4,2)} is symmetric since both (2,4) and (4,2) are present, but it is not transitive because (2,4)&(4,2) would require (2,2), which is absent — 'only symmetric' (III). Relation C = {(1,2),(2,3),(3,4)} has no pair whose reverse is also present, so it is (vacuously) antisymmetric, but (1,2)&(2,3) would require (1,3), which is missing, so it is not transitive — 'antisymmetric' (I). Relation D = {(1,3),(1,4),(2,3),(2,4),(3,1),(3,4)} contains (1,3) and (3,1) with 1≠3 (so not antisymmetric, and also not symmetric since (1,4) lacks its reverse), and (1,3)&(3,1) would require (1,1), which is absent, so it is not transitive either — 'neither symmetric nor transitive' (IV). The match is A-II, B-III, C-I, D-IV.