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(B) The student can select at most $n$ books from $(2n + 1)$ books.Let $T$ be the total number of ways to select at least one book.$T = ^{2n+1}C_1 + ^

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The student can select at most $n$ books from $(2n + 1)$ books.Let $T$ be the total number of ways to select at least one book.$T = ^{2n+1}C_1 + ^{2n+1}C_2 + \dots + ^{2n+1}C_n = 63$.We know that the sum of binomial coefficients is given by:$^{2n+1}C_0 + ^{2n+1}C_1 + \dots + ^{2n+1}C_n + ^{2n+1}C_{n+1} + \dots + ^{2n+1}C_{2n+1} = 2^{2n+1}$.Using the property $^{n}C_r = ^{n}C_{n-r}$,we have $^{2n+1}C_0 = ^{2n+1}C_{2n+1} = 1$ and $^{2n+1}C_1 = ^{2n+1}C_{2n}$,etc.Thus,$^{2n+1}C_0 + 2(^{2n+1}C_1 + ^{2n+1}C_2 + \dots + ^{2n+1}C_n) = 2^{2n+1}$.$1 + 2(T) = 2^{2n+1}$.$1 + 2(63) = 2^{2n+1}$.$1 + 126 = 2^{2n+1} \Rightarrow 127 = 2^{2n+1}$.Wait,re-evaluating the problem statement: If the student selects at most $n$ books,the number of ways is $\sum_{k=1}^{n} {^{2n+1}C_k} = 63$.Since $\sum_{k=0}^{2n+1} {^{2n+1}C_k} = 2^{2n+1}$,and $\sum_{k=0}^{n} {^{2n+1}C_k} = \sum_{k=n+1}^{2n+1} {^{2n+1}C_k} = \frac{1}{2} 	imes 2^{2n+1} = 2^{2n}$.So,$\sum_{k=1}^{n} {^{2n+1}C_k} = 2^{2n} - ^{2n+1}C_0 = 2^{2n} - 1$.Given $2^{2n} - 1 = 63$ $\Rightarrow 2^{2n} = 64$ $\Rightarrow 2^{2n} = 2^6$.$2n = 6 \Rightarrow n = 3$.

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

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