finding largest number of candidate keys that a relation has?

Question

I am trying to solve this question which has to do with candidate keys in a relation. This is the question:

    Consider table R with attributes A, B, C, D, and E. What is the largest number of
candidate keys that R could simultaneously have?

the answer is 10 but i have no clue how it was done, nor how does the word simultaneously plays into effect when calculating the answer.

Answer 1

Sets that are not subsets of other sets.
For example {A-B} and {A,B,C} can't be candidates keys simultaneously, because {A,B} is a subset of {A,B,C}.
Combinations of 2 attributes or 3 attributes generates the maximum number of simultaneous candidates keys.
See how the 3 attributes sets are actually complements of the 2 attributes sets, e.g. {C,D,E} is the complement of {A,B}.

         2               3    
     attributes      attributes
       sets            sets

   1.  {A,B}    -     {C,D,E}
   2.  {A,C}    -     {B,D,E}
   3.  {A,D}    -     {B,C,E}
   4.  {A,E}    -     {B,C,D}
                -     
   5.  {B,C}    -     {A,D,E}
   6.  {B,D}    -     {A,C,E}
   7.  {B,E}    -     {A,C,D}
                -     
   8.  {C,D}    -     {A,B,E}
   9.  {C,E}    -     {A,B,D}
                -     
   10. {D,E}    -     {A,B,C}

---

If I would take sets of a single attribute I would have only 4 options

{A},{B},{C},{D}

Any set with more than 1 element will contain one of the above and therefore will not be qualified.

If I would take sets of 4 attributes I would have only 4 options

{A,B,C,D},{A,B,C,E},{A,B,D,E},{B,C,D,E}

Any set with more than 4 element will contain one of the above and therefore will not be qualified. Any set with less than 4 element will be contained by one of the above and therefore will not be qualified.

etc.