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(A) The total number of ways to select at least one and at most $n$ coins from $(2n + 1)$ distinct coins is given by $T = {}^{2n+1}C_1 + {}^{2n+1}C_2

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The total number of ways to select at least one and at most $n$ coins from $(2n + 1)$ distinct coins is given by $T = {}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_n = 255$.We know that the sum of binomial coefficients is ${}^{2n+1}C_0 + {}^{2n+1}C_1 + \dots + {}^{2n+1}C_{2n+1} = 2^{2n+1}$.Since ${}^{2n+1}C_r = {}^{2n+1}C_{2n+1-r}$,we have ${}^{2n+1}C_0 = {}^{2n+1}C_{2n+1} = 1$.Also,${}^{2n+1}C_1 = {}^{2n+1}C_{2n}$,${}^{2n+1}C_2 = {}^{2n+1}C_{2n-1}$,and so on.Thus,$2({}^{2n+1}C_1 + {}^{2n+1}C_2 + \dots + {}^{2n+1}C_n) + {}^{2n+1}C_0 + {}^{2n+1}C_{2n+1} = 2^{2n+1}$.Substituting $T = 255$,we get $2(255) + 1 + 1 = 2^{2n+1}$.$510 + 2 = 2^{2n+1} \Rightarrow 512 = 2^{2n+1}$.Since $512 = 2^9$,we have $2n + 1 = 9$.$2n = 8 \Rightarrow n = 4$.

$A$ 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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