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(B) The student can select at least one book in $T$ ways,where $T = {}^{2n+1}C_1 + {}^{2n+1}C_2 + ... + {}^{2n+1}C_n = 63$.We know that the sum of all

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$3$

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(B) The student can select at least one book in $T$ ways,where $T = {}^{2n+1}C_1 + {}^{2n+1}C_2 + ... + {}^{2n+1}C_n = 63$.We know that the sum of all binomial coefficients for $(2n+1)$ is ${}^{2n+1}C_0 + {}^{2n+1}C_1 + ... + {}^{2n+1}C_{2n+1} = 2^{2n+1}$.Since ${}^{2n+1}C_r = {}^{2n+1}C_{2n+1-r}$,we have ${}^{2n+1}C_0 = {}^{2n+1}C_{2n+1} = 1$.Thus,$1 + ({}^{2n+1}C_1 + ... + {}^{2n+1}C_n) + ({}^{2n+1}C_{n+1} + ... + {}^{2n+1}C_{2n}) + 1 = 2^{2n+1}$.Since the sum of the first $n$ terms is $T$,and the sum of the next $n$ terms is also $T$,we have $1 + T + T + 1 = 2^{2n+1}$.$2 + 2T = 2^{2n+1} \Rightarrow 1 + T = 2^{2n}$.Given $T = 63$,we get $1 + 63 = 2^{2n} \Rightarrow 64 = 2^{2n}$.$2^6 = 2^{2n}$ $\Rightarrow 2n = 6$ $\Rightarrow n = 3$.

$A$ student is allowed to select at most $n$ books from a collection of $(2n + 1)$ books. If the total number of ways in which he can select at least one book is $63$,then the value of $n$ is

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