How many subsets containing at most n elements from the set of (2n+1) elements can be selected?

Options

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

Answer

To solve the problem of how many subsets containing at most \( n \) elements can be selected from a set of \( 2n + 1 \) elements, we can follow these steps:

### Step 1: Understand the Total Number of Subsets
The total number of subsets of a set with \( m \) elements is given by \( 2^m \). In our case, we have \( 2n + 1 \) elements, so the total number of subsets is:
\[ 2^{2n + 1} \]

### Step 2: Count Subsets with More than \( n \) Elements
To find the number of subsets containing at most \( n \) elements, we can first calculate the number of subsets containing more than \( n \) elements. The subsets with more than \( n \) elements are those that contain \( n + 1 \) to \( 2n + 1 \) elements.

### Step 3: Use the Complement Principle
The number of subsets with more than \( n \) elements can be calculated as:
\[ \text{Number of subsets with more than } n \text{ elements} = \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} \]
Using the property of combinations, we know:
\[ \binom{2n+1}{k} = \binom{2n+1}{2n+1-k} \]
Thus, the number of subsets with more than \( n \) elements is equal to the number of subsets with at most \( n \) elements:
\[ \sum_{k=0}^{n} \binom{2n+1}{k} \]

### Step 4: Relate the Two Counts
Since the total number of subsets is the sum of subsets with at most \( n \) elements and those with more than \( n \) elements, we can write:
\[ 2^{2n + 1} = \sum_{k=0}^{n} \binom{2n+1}{k} + \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} \]
But since the two sums on the right are equal, we can denote:
\[ \sum_{k=0}^{n} \binom{2n+1}{k} = \sum_{k=n+1}^{2n+1} \binom{2n+1}{k} \]
Let \( S = \sum_{k=0}^{n} \binom{2n+1}{k} \). Then we have:
\[ 2^{2n + 1} = S + S = 2S \]
So,
\[ S = \frac{2^{2n + 1}}{2} = 2^{2n} \]

### Conclusion
The number of subsets containing at most \( n \) elements from a set of \( 2n + 1 \) elements is:
\[ \boxed{2^{2n}} \]