Combinatorics: Selection of Candidates

Number of ways to select $1$ to $n$ candidates from $2n+1$ is: $\binom{2n+1}{1} + \binom{2n+1}{2} + \dots + \binom{2n+1}{n} = 63$. By symmetry, $\sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}$. Also $\binom{2n+1}{0} + \sum_{k=1}^n \binom{2n+1}{k} + \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} = 2^{2n+1}$. Since $\sum_{k=1}^n \binom{2n+1}{k} = \sum_{k=n+1}^{2n} \binom{2n+1}{k}$, we have $1 + 2(63) + 1 = 2^{2n+1} \implies 128 = 2^{2n+1} \implies 2n+1 = 7 \implies n = 3$.