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

Hint:

Use symmetry of binomial coefficients: \(\binom{N}{k} = \binom{N}{N-k}\).

Answer 1

The Correct Option is A

Step 1: Formula / Definition}
\[ \sum_{k=1}^n \binom{2n+1}{k} = 255 \]
Step 2: Calculation / Simplification}
\[ \sum_{k=0}^{2n+1} \binom{2n+1}{k} = 2^{2n+1} \]
\[ \binom{2n+1}{0} + \sum_{k=1}^n \binom{2n+1}{k} + \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} = 2^{2n+1} \]
By symmetry, \(\sum_{k=1}^n \binom{2n+1}{k} = \sum_{k=n+1}^{2n} \binom{2n+1}{k}\)
\[ 1 + 255 + 255 + 1 = 2^{2n+1} \Rightarrow 512 = 2^{2n+1} \Rightarrow 2n+1 = 9 \Rightarrow n = 4 \]
Step 3: Final Answer
\[ n = 4 \]