A set contsins `(2n+1)` elements. The number of subset of the set which contain at most n elements is

Options

A. `2^(n)`
B. `2^(n+1)`
C. `2^(n-1)`
D. `2^(2n)`

Answer

To solve the problem of finding the number of subsets of a set containing \(2n + 1\) elements that contain at most \(n\) elements, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Set**:
We have a set with \(2n + 1\) elements. The total number of subsets of any set with \(m\) elements is given by \(2^m\). Therefore, the total number of subsets of our set is:
\[ 2^{2n + 1} \]

2. **Subsets with At Most n Elements**:
We want to find the number of subsets that contain at most \(n\) elements. This means we need to consider subsets with \(0\) elements, \(1\) element, \(2\) elements, up to \(n\) elements.

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

4. **Using the Binomial Theorem**:
According to the binomial theorem, the sum of the binomial coefficients for a given \(m\) is:
\[ \sum_{k=0}^{m} \binom{m}{k} = 2^m \]
Hence, we can also express the sum of the binomial coefficients from \(0\) to \(2n + 1\):
\[ \sum_{k=0}^{2n + 1} \binom{2n + 1}{k} = 2^{2n + 1} \]

5. **Using Symmetry in Binomial Coefficients**:
The binomial coefficients have a symmetry property:
\[ \binom{m}{k} = \binom{m}{m-k} \]
This means that:
\[ \sum_{k=0}^{n} \binom{2n + 1}{k} = \sum_{k=n+1}^{2n + 1} \binom{2n + 1}{k} \]
Therefore, we can conclude that:
\[ \sum_{k=0}^{n} \binom{2n + 1}{k} = \frac{1}{2} \sum_{k=0}^{2n + 1} \binom{2n + 1}{k} = \frac{1}{2} \cdot 2^{2n + 1} = 2^{2n} \]

6. **Final Result**:
Thus, the number of subsets of the set containing at most \(n\) elements is:
\[ 2^{2n} \]

### Final Answer:
The number of subsets of the set which contain at most \(n\) elements is \(2^{2n}\).

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