Candidate Keys

Question

Consider the following relation $R(A, B, C, D, E, F, G, H)$. The maximum number of super keys possible if $4$ simple candidate keys in relation $R$ are ________.

Solution.

What i did is $^8C_4 + ^8C_5 + ^8C_6 + ^8C_7 + ^8C_8 = 163$        Is it wrong ? and why 

Comments

(4C1+4C2+4C3+4C4)×2^(8-4)= 240

(No. of ways for choosing one or two or three or four simple candidate key attributes)×(each of remaining four has two choice)

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@Verma Ashish . I am confused about the terminology.

Can you please elaborate what does the question want us to find?

Answer 1

Since All are Simple Candidate keys..Apply Complementary method. Let's assume $A,B,C,D$ are the Four simple Candidate keys.

All Super Keys = All Possible Combinations -  All combinations Where None of the candidate keys are selected

All Possible Combinations = $2^8$

All combinations Where None of the candidate keys (Say A,B,C,D) are selected :  $2^4$

Hence Our answer = 256 - 16 = 240

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Where you did wrong : 

See, in $_{4}^{8}\textrm{C}$ there is One combination where None of the Candidate keys is selected and All the four selected attributes are Non-key attribute. 

Further, Even if you choose Only One attribute from the $8$ But which is a candidate key (A or B or C or D), It is a Super key. And Such Combinations are not counted in your answer.

So, To Correct your answer, you would have to add such combinations. It is gonna be Time-taking approach (But you can try for learning purpose), That's why In This Question I think Complementary approach is Better one.

Comments

@Deepakk Poonia (Dee) Sir, In this 

All Super Keys = All Possible Combinations -  All combinations Where None of the candidate keys are selected

The case where Key is Φ ( None of them are taken) also comes under “ All possible combinations ”.

But is removed because it also comes under “ All combinations where none of candidate keys are selected “ part , so subtraction gives us the super keys that are made only including our simple candidate keys ,

Am I right?

I got this doubt, hence I wrote here so anyone having same doubt as me will understand !

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In the solution, as we have considered A,B,C,D as candidate keys, Why don’t we need to multiply 8C4 for choosing such possible combinations of candidate keys?