Answer

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Hint – In this question we have to select at least (that is minimum) one coin and at most n coins (that is maximum). The total number of coins given is (2n+1). Use the concept that the ways of selecting r coins out of n coins is $^n{C_r}$. Keep in mind that there is atleast an atmost concept that will be used in the question.

Complete step-by-step answer:

Number of distinct coins are = (2n + 1)

We have to find out the value of n.

Now a person is permitted to select at least one and at most n coins and the total number of ways in which he can select coins is 255.

Number of ways of selecting 1 coin from (2n + 1) coins is ${}^{2n + 1}{C_1}$

Similarly Number of ways of selecting 2 coin from (2n + 1) coins is ${}^{2n + 1}{C_2}$

Therefore Number of ways of selecting n coin from (2n + 1) coins is ${}^{2n + 1}{C_n}$

The total number of ways in which he can select coins is 255.

$ \Rightarrow {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2} + ............ + {}^{2n + 1}{C_n} = 255$……………………………. (1)

Now according to binomial theorem the expansion of ${\left( {1 + x} \right)^{2n + 1}}$ is

${\left( {1 + x} \right)^{2n + 1}} = {}^{2n + 1}{C_0} + {}^{2n + 1}{C_1}x + .......... + {}^{2n + 1}{C_n}{x^n} + {}^{2n + 1}{C_{n + 1}}{x^{n + 1}} + ....................... + {}^{2n + 1}{C_{2n}}{x^{2n}} + {}^{2n + 1}{C_{2n + 1}}{x^{2n + 1}}$

Now in above equation put x = 1 we have,

${\left( {1 + 1} \right)^{2n + 1}} = {}^{2n + 1}{C_0} + {}^{2n + 1}{C_1}.1 + .......... + {}^{2n + 1}{C_n}.1 + {}^{2n + 1}{C_{n + 1}}.1 + ....................... + {}^{2n + 1}{C_{2n}}.1 + {}^{2n + 1}{C_{2n + 1}}.1$

${\left( 2 \right)^{2n + 1}} = {}^{2n + 1}{C_0} + {}^{2n + 1}{C_1} + .......... + {}^{2n + 1}{C_n} + {}^{2n + 1}{C_{n + 1}} + ....................... + {}^{2n + 1}{C_{2n}} + {}^{2n + 1}{C_{2n + 1}}$

Now as we know that the value of ${}^{2n + 1}{C_0} = {}^{2n + 1}{C_{2n + 1}} = 1$ and ${}^n{C_r} = {}^n{C_{n - r}}$

$ \Rightarrow {}^{2n + 1}{C_1} = {}^{2n + 1}{C_{2n + 1 - 1}} = {}^{2n + 1}{C_{2n}}$ So, using this property in above equation we have,

$ \Rightarrow {\left( 2 \right)^{2n + 1}} = 1 + 2\left( {{}^{2n + 1}{C_1} + .......... + {}^{2n + 1}{C_n}} \right) + 1$

Now from equation (1) we have,

$ \Rightarrow {\left( 2 \right)^{2n + 1}} = 1 + 2\left( {255} \right) + 1$

Now simplify the above equation we have,

$ \Rightarrow {2.2^{2n}} = 2 + 2\left( {255} \right)$

Now divide by 2 in above equation we have,

$ \Rightarrow {2^{2n}} = 256 = {2^8}$

So on comparing

$ \Rightarrow 2n = 8$

$ \Rightarrow n = 4.$

Hence option (a) is correct.

Note – Whenever we face such types of problems the key concept is to use the concept of permutation and combination along with the possible case scenarios to satisfy the condition of at most coins drawn and at least coin drawn. This concept will help you get on the right track to reach the answer.

Complete step-by-step answer:

Number of distinct coins are = (2n + 1)

We have to find out the value of n.

Now a person is permitted to select at least one and at most n coins and the total number of ways in which he can select coins is 255.

Number of ways of selecting 1 coin from (2n + 1) coins is ${}^{2n + 1}{C_1}$

Similarly Number of ways of selecting 2 coin from (2n + 1) coins is ${}^{2n + 1}{C_2}$

Therefore Number of ways of selecting n coin from (2n + 1) coins is ${}^{2n + 1}{C_n}$

The total number of ways in which he can select coins is 255.

$ \Rightarrow {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2} + ............ + {}^{2n + 1}{C_n} = 255$……………………………. (1)

Now according to binomial theorem the expansion of ${\left( {1 + x} \right)^{2n + 1}}$ is

${\left( {1 + x} \right)^{2n + 1}} = {}^{2n + 1}{C_0} + {}^{2n + 1}{C_1}x + .......... + {}^{2n + 1}{C_n}{x^n} + {}^{2n + 1}{C_{n + 1}}{x^{n + 1}} + ....................... + {}^{2n + 1}{C_{2n}}{x^{2n}} + {}^{2n + 1}{C_{2n + 1}}{x^{2n + 1}}$

Now in above equation put x = 1 we have,

${\left( {1 + 1} \right)^{2n + 1}} = {}^{2n + 1}{C_0} + {}^{2n + 1}{C_1}.1 + .......... + {}^{2n + 1}{C_n}.1 + {}^{2n + 1}{C_{n + 1}}.1 + ....................... + {}^{2n + 1}{C_{2n}}.1 + {}^{2n + 1}{C_{2n + 1}}.1$

${\left( 2 \right)^{2n + 1}} = {}^{2n + 1}{C_0} + {}^{2n + 1}{C_1} + .......... + {}^{2n + 1}{C_n} + {}^{2n + 1}{C_{n + 1}} + ....................... + {}^{2n + 1}{C_{2n}} + {}^{2n + 1}{C_{2n + 1}}$

Now as we know that the value of ${}^{2n + 1}{C_0} = {}^{2n + 1}{C_{2n + 1}} = 1$ and ${}^n{C_r} = {}^n{C_{n - r}}$

$ \Rightarrow {}^{2n + 1}{C_1} = {}^{2n + 1}{C_{2n + 1 - 1}} = {}^{2n + 1}{C_{2n}}$ So, using this property in above equation we have,

$ \Rightarrow {\left( 2 \right)^{2n + 1}} = 1 + 2\left( {{}^{2n + 1}{C_1} + .......... + {}^{2n + 1}{C_n}} \right) + 1$

Now from equation (1) we have,

$ \Rightarrow {\left( 2 \right)^{2n + 1}} = 1 + 2\left( {255} \right) + 1$

Now simplify the above equation we have,

$ \Rightarrow {2.2^{2n}} = 2 + 2\left( {255} \right)$

Now divide by 2 in above equation we have,

$ \Rightarrow {2^{2n}} = 256 = {2^8}$

So on comparing

$ \Rightarrow 2n = 8$

$ \Rightarrow n = 4.$

Hence option (a) is correct.

Note – Whenever we face such types of problems the key concept is to use the concept of permutation and combination along with the possible case scenarios to satisfy the condition of at most coins drawn and at least coin drawn. This concept will help you get on the right track to reach the answer.

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