0 0 votes For a scholarship, at the most n candidates out of 2n + 1 can be selected. If the number of different ways of selection of at least one candidate is 63, the maximum number of candidates that can be selected for the scholarship is 3 4 6 5 Quantitative Aptitude cat1999 quantitative-aptitude + – go_editor 14.2k points 4.5k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
2 2 votes The number of ways you can select at least 1 candidate up to n candidates out of the total 2n+1 is given as 63. $^{2n+1}C_1+^{2n+1}C_2+...+^{2n+1}C_n$=63 and $^{2n+1}C_0$+$^{2n+1}C_1$+$^{2n+1}C_2$+...+$^{2n+1}C_n$+$^{2n+1}C_n+1$+$^{2n+1}C_n+2$+...+$^{2n+1}C_2n+1$=$2^{2n+1}$ ..............(1) We know $^{2n+1}C_0=1$ and $^n{C}_r = ^{n}C_n-r$ So $^{2n+1}C_0+^{2n+1}C_1+^{2n+1}C_2+...+^{2n+1}C_n=^{2n+1}C_n+1+^{2n+1}Cn+2+...+^{2n+1}C_2n+1$ ...........(2) From(1) and (2) 1+63+63+1=$2^{2n+1}$ $2^{7}$ =$2^{2n+1}$ n=3 Hence,Option(A)3. Leen Sharma answered May 12, 2016 • edited May 12, 2016 by Leen Sharma Leen Sharma 11.6k points comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Answer D) 5 63 different ways we can select at least 1 candidate 2n+1 =63 n=31 Max no of candidate can select log31 =5 srestha answered May 12, 2016 srestha 5.2k points comment Share Follow See 1 comment 1 1 comment reply Leen Sharma 11.6k points commented May 12, 2016 reply Follow flag You did wrong because 2n+1 is total number of student and 63 is number of different ways of selection for scholarship.So they can't be equal. and one more thing how log31 =5 ? 0 0 replyShare Please log in or register to add a comment.