Question

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?

• A. $4$
• B. $8$
• C. $16$
• D. $32$

Answer 1

The correct option is A

$4$

Explanation for the correct option:

Step 1: Express an equation for the given situation:

Given: Number of ways of selecting one to $n$ coins from a collection of $(2n+1)$ distinct coins, is $255.$

In mathematical terms, It can be expressed as follows:

${}^{2n+1}C_1+{}^{2n+1}C_2+....+{}^{2n+1}C_n=255$

Step 2: Find the sum of the obtained expression:

Use Binomial coefficients relation to find the sum.

$\begin{array}{rcl}{}^{2n+1}C_0+{}^{2n+1}C_1+{}^{2n+1}C_2+....+{}^{2n+1}C_n+{}^{2n+1}C_{n+1}.....+{}^{2n+1}C_{2n+1}& =& 2^{2n+1}\\ {}^{2n+1}C_0+2\left\{{}^{2n+1}C_1+{}^{2n+1}C_2+....+{}^{2n+1}C_n\right\}+{}^{2n+1}C_{2n+1}& =& 2^{2n+1}\\ 1+2\left\{{}^{2n+1}C_1+{}^{2n+1}C_2+....+{}^{2n+1}C_n\right\}+1& =& 2^{2n+1}\\ 2+2\left\{{}^{2n+1}C_1+{}^{2n+1}C_2+....+{}^{2n+1}C_n\right\}& =& 2^{2n+1}\end{array}$

Step 3: Put the given value and simplify the equation:

Put the given value of the sum in the obtained expression.

$\begin{array}{rcl}2+2\left\{255\right\}& =& 2^{2n+1}\\ 255+1& =& 2^{2n}\\ 2^{2n}& =& 2^8\\ n& =& 4\end{array}$

Hence, option (A) is the correct answer.