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(A) Let the total number of ways to select at most $n$ books be $x$. Since the student must select at least one book,we have:$x = {}^{2n+1}C_1 + {}^{2

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$4$

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(A) Let the total number of ways to select at most $n$ books be $x$. Since the student must select at least one book,we have:$x = {}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_n = 255$We know that the sum of all combinations for $2n+1$ items is:${}^{2n+1}C_0 + {}^{2n+1}C_1 + \dots + {}^{2n+1}C_n + {}^{2n+1}C_{n+1} + \dots + {}^{2n+1}C_{2n+1} = 2^{2n+1}$Using the property ${}^{m}C_r = {}^{m}C_{m-r}$,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:$2({}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_n) + {}^{2n+1}C_0 + {}^{2n+1}C_{2n+1} = 2^{2n+1}$$2x + 1 + 1 = 2^{2n+1}$$2x + 2 = 2^{2n+1}$$x + 1 = 2^{2n}$Given $x = 255$,we get:$255 + 1 = 2^{2n}$$256 = 2^{2n}$$2^8 = 2^{2n}$$2n = 8 \implies n = 4$Therefore,the value of $n$ is $4$.

$A$ student is allowed to choose 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 $255$,then the value of $n$ is:

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