A person is permitted to select at least one | vedclass

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Question

(a) Since the person is allowed to select at most $n$ coins out of $(2n + 1)$ coins,

therefore in order to select one, two, three, &hellip;., $n$ c

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(a) Since the person is allowed to select at most $n$ coins out of $(2n + 1)$ coins,

therefore in order to select one, two, three, &hellip;., $n$ coins.

Thus, if T is the total number of ways of selecting one coin, then

$T = {\,^{2n + 1}}{C_1}\,{ + ^{2n + 1}}{C_2}\, + \,...... + {\,^{2n + 1}}{C_n}\, = 255$&hellip;..$(i)$

Again the sum of binomial coefficients

${ = ^{2n + 1}}{C_0}{\mkern 1mu} { + ^{2n + 1}}{C_1}{\mkern 1mu}  + {{\mkern 1mu} ^{2n + 1}}{C_2} + ..... + $

$^{2n + 1}{C_n}{\mkern 1mu} { + ^{2n + 1}}{C_{n + 1}} + {{\mkern 1mu} ^{2n + 1}}{C_{n + 2}}{\mkern 1mu}  + ..... + $

$^{2n + 1}{C_{2n + 1}} = {(1 + 1)^{2n + 1}} = {2^{2n + 1}}$

$ =  = { > ^{2n + 1}}{C_0}{\mkern 1mu}  + {\mkern 1mu} 2{\rm{(}}2n + 1{C_1}{\mkern 1mu}  + {{\mkern 1mu} ^{2n + 1}}{C_2}{\mkern 1mu}  + ...{ + ^{2n + 1}}{C_n}{\rm{)}}$

$ + {{\mkern 1mu} ^{2n + 1}}{C_{2n + 1}}{\mkern 1mu}  = {2^{2n + 1}}$

==> $1 + 2(T) + 1 = {2^{2n + 1}} \Rightarrow 1 + T = \frac{{{2^{2n + 1}}}}{2} = {2^{2n}}$

==> $1 + 255 = {2^{2n}}\, \Rightarrow \,\,{2^{2n}}\,\, = \,\,{2^8}\, \Rightarrow \,\,n\,\, = \,\,4$.

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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