Let `A = {1,2,3,4,...,n}` be a set containing n elements, then for any given `k leqn`, the number of subsets of A having k as largest element must be

Options

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

Explanation

To solve the problem, we need to find the number of subsets of the set \( A = \{1, 2, 3, \ldots, n\} \) that have \( k \) as the largest element.

### Step-by-Step Solution:

1. **Understanding the Set**:
The set \( A \) contains \( n \) elements: \( A = \{1, 2, 3, \ldots, n\} \).

2. **Condition on \( k \)**:
We are given that \( k \leq n \). This means \( k \) can be any element from the set \( A \).

3. **Identifying Elements in the Subset**:
For a subset to have \( k \) as the largest element, all other elements in the subset must be from the set \( \{1, 2, 3, \ldots, k-1\} \). This is because any element larger than \( k \) would violate the condition of \( k \) being the largest.

4. **Counting Eligible Elements**:
The elements that can be included in the subset, apart from \( k \), are \( \{1, 2, 3, \ldots, k-1\} \). There are \( k-1 \) such elements.

5. **Forming Subsets**:
Each of the \( k-1 \) elements can either be included in the subset or not. Therefore, for each of these elements, there are 2 choices (include or exclude).

6. **Calculating Total Subsets**:
The total number of subsets that can be formed from \( k-1 \) elements is given by \( 2^{k-1} \). This includes the empty subset as well.

7. **Conclusion**:
Thus, the number of subsets of \( A \) that have \( k \) as the largest element is \( 2^{k-1} \).

### Final Answer:
The number of subsets of \( A \) having \( k \) as the largest element is \( 2^{k-1} \).

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