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(A) The total number of ways to select $r$ coins from $(2n + 1)$ distinct coins is given by $\binom{2n+1}{r}$.Given that the person selects at least $

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$4$

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(A) The total number of ways to select $r$ coins from $(2n + 1)$ distinct coins is given by $\binom{2n+1}{r}$.Given that the person selects at least $1$ and at most $n$ coins,the total number of ways $T$ is:$T = \binom{2n+1}{1} + \binom{2n+1}{2} + \dots + \binom{2n+1}{n} = 255$.We know that the sum of binomial coefficients is $\sum_{r=0}^{2n+1} \binom{2n+1}{r} = 2^{2n+1}$.Since $\binom{2n+1}{r} = \binom{2n+1}{2n+1-r}$,we have $\binom{2n+1}{0} = \binom{2n+1}{2n+1} = 1$.Thus,$\binom{2n+1}{0} + \binom{2n+1}{1} + \dots + \binom{2n+1}{n} + \binom{2n+1}{n+1} + \dots + \binom{2n+1}{2n+1} = 2^{2n+1}$.This simplifies to $1 + T + T + 1 = 2^{2n+1}$,which is $2 + 2T = 2^{2n+1}$.Dividing by $2$,we get $1 + T = 2^{2n}$.Substituting $T = 255$,we get $1 + 255 = 2^{2n}$,so $256 = 2^{2n}$.Since $256 = 2^8$,we have $2n = 8$,which implies $n = 4$.

$A$ person is permitted to select at least one and at most $n$ coins from a collection of $(2n + 1)$ distinct coins. If the total number of ways in which he can select coins is $255$,then $n$ equals

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