A set contains (2n+1) elements . IF the number of subsets of this set which contain at most n elements is 4096, then n is equal to

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem, we need to find the value of \( n \) given that the number of subsets of a set containing \( (2n + 1) \) elements, which contain at most \( n \) elements, is equal to 4096.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a set with \( 2n + 1 \) elements. We need to find subsets that contain at most \( n \) elements.

2. **Using the Binomial Coefficient**:
The number of ways to choose \( r \) elements from a set of \( m \) elements is given by the binomial coefficient \( \binom{m}{r} \). Therefore, the total number of subsets with at most \( n \) elements can be expressed as:
\[ \sum_{r=0}^{n} \binom{2n+1}{r} \]

3. **Using the Binomial Theorem**:
According to the binomial theorem, the sum of the binomial coefficients from \( 0 \) to \( m \) is equal to \( 2^m \):
\[ \sum_{r=0}^{2n+1} \binom{2n+1}{r} = 2^{2n+1} \]
Since the binomial coefficients are symmetric, we have:
\[ \sum_{r=0}^{n} \binom{2n+1}{r} = \sum_{r=n+1}^{2n+1} \binom{2n+1}{r} \]
Therefore, we can say:
\[ \sum_{r=0}^{n} \binom{2n+1}{r} = \frac{1}{2} \cdot 2^{2n+1} = 2^{2n} \]

4. **Setting Up the Equation**:
According to the problem, this sum equals 4096:
\[ 2^{2n} = 4096 \]

5. **Solving for \( n \)**:
We can express 4096 as a power of 2:
\[ 4096 = 2^{12} \]
Therefore, we have:
\[ 2^{2n} = 2^{12} \]
This implies:
\[ 2n = 12 \]
Dividing both sides by 2 gives:
\[ n = 6 \]

6. **Conclusion**:
The value of \( n \) is \( 6 \).