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(B) The student is allowed to select at least one book and at most $n$ books out of $(2n + 1)$ books.Thus,the total number of ways $T$ is given by:$T

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(B) The student is allowed to select at least one book and at most $n$ books out of $(2n + 1)$ books.Thus,the total number of ways $T$ is given by:$T = {^{2n+1}C_1} + {^{2n+1}C_2} + ... + {^{2n+1}C_n} = 63$ ..... $(i)$We know that the sum of binomial coefficients is:${^{2n+1}C_0} + {^{2n+1}C_1} + ... + {^{2n+1}C_n} + {^{2n+1}C_{n+1}} + ... + {^{2n+1}C_{2n+1}} = 2^{2n+1}$Using the property ${^mC_r} = {^mC_{m-r}}$,we have ${^{2n+1}C_0} = {^{2n+1}C_{2n+1}} = 1$ and ${^{2n+1}C_1} = {^{2n+1}C_{2n}}$,etc.Thus,the sum can be written as:$1 + ({^{2n+1}C_1} + ... + {^{2n+1}C_n}) + ({^{2n+1}C_{n+1}} + ... + {^{2n+1}C_{2n+1}}) = 2^{2n+1}$Since ${^{2n+1}C_1} + ... + {^{2n+1}C_n} = 63$ and by symmetry,${^{2n+1}C_{n+1}} + ... + {^{2n+1}C_{2n+1}} = 63$,we get:$1 + 63 + 63 = 2^{2n+1}$$127 = 2^{2n+1}$Wait,re-evaluating the logic: The sum of the first $(n+1)$ terms is $2^{2n+1} / 2 = 2^{2n}$.So,${^{2n+1}C_0} + {^{2n+1}C_1} + ... + {^{2n+1}C_n} = 2^{2n}$.$1 + 63 = 2^{2n} \Rightarrow 64 = 2^{2n} \Rightarrow 2^6 = 2^{2n}$.Therefore,$2n = 6$,which gives $n = 3$.

$A$ 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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