# 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 in which he can select coins is 255, then n equals

$

(a){\text{ 4}} \\

(b){\text{ 8}} \\

(c){\text{ 16}} \\

(d){\text{ 32}} \\

$

Last updated date: 17th Mar 2023

•

Total views: 303.6k

•

Views today: 5.82k

Answer

Verified

303.6k+ views

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.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?