A relation RABCDEF has one and only one candidate key How ma
A relation R(A,B,C,D,E,F) has one and only one candidate key. How many superkeys can there be?
Database class.
Solution
total superkeys = 2^(n-2)
n(A+B+C+D+E+F)= n(A) + n(B) + n(C) + n(D) + n(E) + n(F) - 2(ABCDEF)
So total superkeys --> 2^(n-1) + 2^(n-1) + 2^(n-1) + 2^(n-1) + 2^(n-1) + 2^(n-1) - 2^(n-6)
