Question:
A student is allowed to select at most $n$ books from a collection of $2n+1$ books. If the total number of ways in which he can select at least one book is $255$, then the value of $n$ is
A student is allowed to select at most $n$ books from a collection of $2n+1$ books. If the total number of ways in which he can select at least one book is $255$, then the value of $n$ is
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Remember: For odd $m$, the sum of the first half of the binomial coefficients equals $2^{m-1}$.
Updated On: Jun 3, 2026
- $6$
- $5$
- $4$
- $3$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
Use symmetry of binomial coefficients.
Step 2: Meaning
The number of ways of selecting at most $n$ books from $2n+1$ books is \[ \sum_{r=1}^{n}\binom{2n+1}{r}. \] Since $2n+1$ is odd, \[ \sum_{r=0}^{n}\binom{2n+1}{r} = 2^{2n}. \]
Step 3: Analysis
Therefore, \[ \sum_{r=1}^{n}\binom{2n+1}{r} = 2^{2n}-1. \] Given \[ 2^{2n}-1=255. \] Hence, \[ 2^{2n}=256=2^8. \] Thus, \[ 2n=8 \quad\Rightarrow\quad n=4. \]
Step 4: Conclusion
Therefore the required value is $4$.
Final Answer: (C)
Use symmetry of binomial coefficients.
Step 2: Meaning
The number of ways of selecting at most $n$ books from $2n+1$ books is \[ \sum_{r=1}^{n}\binom{2n+1}{r}. \] Since $2n+1$ is odd, \[ \sum_{r=0}^{n}\binom{2n+1}{r} = 2^{2n}. \]
Step 3: Analysis
Therefore, \[ \sum_{r=1}^{n}\binom{2n+1}{r} = 2^{2n}-1. \] Given \[ 2^{2n}-1=255. \] Hence, \[ 2^{2n}=256=2^8. \] Thus, \[ 2n=8 \quad\Rightarrow\quad n=4. \]
Step 4: Conclusion
Therefore the required value is $4$.
Final Answer: (C)
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