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(A) The number of ways to select at most $n$ books from $(2n+1)$ books is given by the sum of combinations:${}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{

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$4$

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(A) The number of ways to select at most $n$ books from $(2n+1)$ books is given by the sum of combinations:${}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_n = 255$ $(i)$Using the property ${}^mC_r = {}^mC_{m-r}$,we have:${}^{2n+1}C_{2n} + {}^{2n+1}C_{2n-1} + \dots + {}^{2n+1}C_{n+1} = 255$ $(ii)$Adding $(i)$ and $(ii)$:$({}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_{2n}) = 510$Adding ${}^{2n+1}C_0$ and ${}^{2n+1}C_{2n+1}$ (both equal to $1$) to both sides:${}^{2n+1}C_0 + {}^{2n+1}C_1 + \dots + {}^{2n+1}C_{2n+1} = 510 + 1 + 1 = 512$Since the sum of binomial coefficients $\sum_{k=0}^{m} {}^mC_k = 2^m$,we have:$2^{2n+1} = 512 = 2^9$Equating the exponents:$2n + 1 = 9$$2n = 8$$n = 4$

$A$ student is allowed to select at most $n$ books from a collection of $(2n+1)$ books. If the total number of ways in which one can select at least one book is $255$,then the value of $n$ is

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