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

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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}

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

This is the exact formula.

Source: Made easy test series

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