Given STUDENT(Rollno, Name, Courseno) and COURSE(Courseno, Coursename, Capacity) with 5 and 3 records, R = STUDENT × COURSE is computed. Match List I with List II. List I: A. Degree of table R, B. Cardinality of table R, C. Foreign key of relation STUDENT, D. Foreign key of relation COURSE. List II: I. 15, II. NIL, III. 6, IV. Courseno.

1. AA ‐III , B ‐I , C ‐IV , D ‐ II
2. BA ‐I , B ‐III , C ‐IV , D ‐ II
3. CA ‐I , B ‐III , C ‐IV , D ‐ II
4. DA ‐I , B ‐III, C ‐II , D ‐ IV

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Correct answer: A. A ‐III , B ‐I , C ‐IV , D ‐ II

Explanation

Degree = 3 + 3 = 6 attributes; cardinality = 5 × 3 = 15 tuples; STUDENT.Courseno references COURSE; COURSE has no foreign key.