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.