ANSWER IS 21
Candidate Keys are AD AE BD BE CD CE.
Now, we have to find out Super keys. Let's consider the first candidate key(CK);
1st CK= AD {_ _ _} we can fill 3 space with B, C, E so combinations will be 2^3 = 8
8 Super keys from first CK.
2nd CK= AE{_ _ _} B C D
Now, the important point here is, in the first candidate key attribute A with attribute D
and in Second candidate key, we have attribute A (So, the combination of AE with attribute D we can eliminate directly because those combinations we already considered in first CK).
so combinations will be 2^2 = 4
4 Super keys from 2nd CK.
3rd CK= BD{_ _ _} A C E
Similarly here, in the first candidate key attribute A with attribute D
So, the combination of BD with the attribute A we can eliminate directly because those combinations we already considered in first CK).
so combinations will be 2^2 = 4
4 Super keys from 3rd CK.
4th CK= BE{_ _ _} A C D
Similarly here, in the 2nd and 3rd candidate key's attributes A and D with attribute B
So, the combination of BE with the attributes A and D we can eliminate directly because those combinations we already considered in 2nd and 3rd CKs).
so combinations will be 2^1 = 2
2 Super keys from 4th CK.
Similarly for CD, 2 Super keys from 5th CK
and 1 Super key from 6th CK.
Total Super keys= 8+4+4+2+2+1= 21