candidate key normalization

Question

What is the maximum number of possible candidate key for relation on n attributes.

Answer 1

If a table has n attributes, max candidate keys possible = $nC \left ( n/2 \right )$
e.g. for relation R(ABCD) max canndidate keys = 4C2 = 6
CK = {AB, AC, AD, BC, BD, CD}

Comments

ABC becomes super key not a candidate key when you consider AC as candidate key

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Correct! Answer edited.

Answer 2

Ans is nC(n/2)

When you group the attributes for candidate keys you will get maximum number when they are grouped by n/2 element's

Answer 3

$^{n}C_{\left \lfloor \frac{n}{2} \right \rfloor}$

This is the exact formula.

Source: Made easy test series

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Comments

Here ceil and floor do not make any difference.  so simply NC(N/2) would be enough.

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yes it is correct because 5c2=5c3   due to ncr=nc(n-r)

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yes it is correct because 5c2=5c3   due to ncr=nc(n-r)