Q1) Consider the relation with n attributes if any 2 attribute is a candidate key, then the total number of candidate keys are___ ?
Then what will be the generalized formula
I) $\sum_{i=2}^{n}C(n,i)$
II) $\sum_{i=2}^{n}C(n,i)*C(i,2)$
I think second one should be the formula for it.
Suppose we have R(ABCD) and any two attribute is a candidate key, then two different candidate keys with different extra attribute will not be calculated as single super key.
For example, we have the Super key ABE with different candidate keys(mentioned in bold letters) and with different extra attributes which are ABE, ABE, and AEB
Above is superkeys are according to the second formula but if we go according to the first formula then ABE is a single superkey.
So which one is correct?
Q2) Is the answer remains same if use every in place of any in above question?
