Here candidate keys are already given .So we need to use only set theory to find no of superkeys :
a) Candidate key one at a time :
(i) Due to AB : Now we can either add or not add each of attributes C,D and E to AB to form superkey as we know candidate key is a minimal superkey..So for each of the 3 attributes , we have 3 choices
So no of superkeys due to AB = 23 = 8
Similarly no of superkeys due to BC = 8
Similarly no of superkeys due to CD = 8
b) Candidate key 2 at a time :
(i) Due to AB,BC : We have 2 attributes remaining D and E..So no of ways = 22 = 4
(ii) Due to BC,CD : We have 2 attributes remaining A and E..So no of ways = 22 = 4
(iii) Due to AB,CD : We have only E remaining so no of ways = 2
c) Candidate key taken all at a time :
i.e. due to AB,BC and CD : We have only E remaining so no of ways = 2
Now having done this we apply inclusion exclusion principle :
No of superkeys = Superkeys due to candidate key one at a time
- Superkeys due to candidate key one at a time
+ Superkeys due to candidate key one at a time
= 24 - 10 + 2
= 16
Therefore no of superkeys = 16..Hope this clears your query..
