Q69 / 200 · Manipal (MET) 2013 · Mathematics
A person is permitted to select at least one and at most $n$ coins from a collection of $(2 n+1)$ distinct coins. It the total number of ways in which he can select coins is 255 , then $n$ equals
- Correct answerA$4$
- B$8$
- C$16$
- D$32$
Solution
We have
${ }^{2 n+1} C_1+{ }^{2 n+1} C_2+\ldots+{ }^{2 n+1} C_n=255$
Also, the sum of binomial coefficients
$\begin{aligned} & { }^{2 n+1} C_0+{ }^{2 n+1} C_1+{ }^{2 n+1} C_2+\ldots+{ }^{2 n+1} C_n \\ & \quad+{ }^{2 n+1} C_{n+1}++{ }^{2 n+1} C_{n+2}+\ldots+{ }^{2 n+1} C_{2 n+1}\end{aligned}$
$=(1+1)^{2 n+1}=2^{2 n+1}$
$\begin{array}{r}\Rightarrow{ }^{2 n+1} C_0+2\left\{{ }^{2 n+1} C_1+{ }^{2 n+1} C_2+\ldots+{ }^{2 n+1} C_n\right\} \\ +{ }^{2 n+1} C_{2 n+1}=2^{2 n+1}\end{array}$
$\begin{array}{rlrl}\Rightarrow & & 1+2(255)+1 & =2^{2 n+1} \\ \Rightarrow & 1+255 & =2^{2 n} \\ \Rightarrow & & 2^{2 n} & =2^8 \\ \Rightarrow & & n & =4\end{array}$
Hence, the correct answer is option: 'A'.