Self Doubt

Question

Consider the following relation:

R (A1, A2, ….An) and every (n-2) attributes of R forms a candidate key. How many super
keys are there in R?

Comments

$\Large \binom{n}{n-2} + \binom{n}{n-1} + 1$

Answer 1

As every n-2 attributes of relation R form a candidate key it implies that every n-1 attributes will also be a super key(because it contains the n-2 attributes that form a candidate key previously)

so total number of super keys:

as every candidate key is a super key :

 $\binom{n}{n-2}$ candidate keys will be there.

taking all possible combinations of 'n-1' attributes together they will also form a super key:

$\binom{n}{n-1}$ 

taking all possible combinations of 'n' attributes together :

$\binom{n}{n}$ (which will be the complete attribute set)

so final answer would be:

 

 $\binom{n}{n-2}$+$\binom{n}{n-1}$ +$\binom{n}{n}$=$\frac{n^{2}+n+2}{2}$