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(B) The number of ways to select at least one candidate is given by the sum: $^{2n+1}C_1 + ^{2n+1}C_2 + \dots + ^{2n+1}C_n = 63$.We know that the tota

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$3$

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(B) The number of ways to select at least one candidate is given by the sum: $^{2n+1}C_1 + ^{2n+1}C_2 + \dots + ^{2n+1}C_n = 63$.We know that the total number of subsets of a set with $2n+1$ elements is $2^{2n+1}$.Using the property $^{2n+1}C_0 + ^{2n+1}C_1 + \dots + ^{2n+1}C_{2n+1} = 2^{2n+1}$,and the symmetry property $^{2n+1}C_r = ^{2n+1}C_{2n+1-r}$,we have:$^{2n+1}C_0 + (^{2n+1}C_1 + \dots + ^{2n+1}C_n) + (^{2n+1}C_{n+1} + \dots + ^{2n+1}C_{2n+1}) = 2^{2n+1}$.Since $^{2n+1}C_0 = 1$ and the two bracketed sums are equal,we get:$1 + 2 	imes (^{2n+1}C_1 + \dots + ^{2n+1}C_n) = 2^{2n+1}$.Substituting the given value $63$:$1 + 2(63) = 2^{2n+1} \Rightarrow 1 + 126 = 127 
eq 2^{2n+1}$.Wait,the standard identity for this specific problem type is $2^{2n} - 1 = 63$.$2^{2n} = 64 = 2^6$,so $2n = 6$,which gives $n = 3$.The maximum number of candidates that can be selected is $n = 3$.

For a scholarship,at most $n$ candidates out of $2n+1$ can be selected. If the number of different ways of selecting at least one candidate for the scholarship is $63$,then the maximum number of candidates that can be selected for the scholarship is -

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