5 votes 5 votes What is the maximum number of possible candidate key for relation on n attributes. Aryan Asrafi asked Sep 3, 2016 Aryan Asrafi 5.5k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
5 votes 5 votes 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} Rakеsh Kumar answered Sep 3, 2016 edited Sep 3, 2016 by Rakеsh Kumar Rakеsh Kumar comment Share Follow See all 2 Comments See all 2 2 Comments reply Pavan Kumar Munnam commented Sep 3, 2016 reply Follow Share ABC becomes super key not a candidate key when you consider AC as candidate key 1 votes 1 votes Rakеsh Kumar commented Sep 3, 2016 reply Follow Share Correct! Answer edited. 0 votes 0 votes Please log in or register to add a comment.
3 votes 3 votes 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 Pavan Kumar Munnam answered Sep 3, 2016 Pavan Kumar Munnam comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes $^{n}C_{\left \lfloor \frac{n}{2} \right \rfloor}$ This is the exact formula. Source: Made easy test series Verified by R a v i n d r a B a b u R a v u l a shashankrustagi answered Oct 28, 2020 shashankrustagi comment Share Follow See all 3 Comments See all 3 3 Comments reply VYAN_jy commented Oct 16, 2021 reply Follow Share Here ceil and floor do not make any difference. so simply NC(N/2) would be enough. 0 votes 0 votes Chandrabhan Vishwa 1 commented Oct 16, 2021 reply Follow Share yes it is correct because 5c2=5c3 due to ncr=nc(n-r) 1 votes 1 votes Chandrabhan Vishwa 1 commented Oct 16, 2021 reply Follow Share yes it is correct because 5c2=5c3 due to ncr=nc(n-r) 0 votes 0 votes Please log in or register to add a comment.