Explanation:
To solve this problem, we need to understand the concept of subsets and combinatorics.

Understanding Subsets:
A subset is a set that contains elements from another set. For example, if we have a set A = {1, 2, 3}, then the subsets of A are {}, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3}. The number of subsets of a set with n elements is 2^n.

Understanding the Problem:
We are given a set with 2n+1 elements. We need to find the number of subsets of this set that contain more than n elements.

Solution:
Let's consider the set S with 2n+1 elements. We can divide this set into two parts: the first part contains n elements and the second part contains n+1 elements.

Case 1: Choosing n+1 elements from the second part
In this case, we need to select n+1 elements from the second part of the set. The number of ways to do this is given by the binomial coefficient (n+1)C(n+1) = 1.

Case 2: Choosing n elements from the first part
In this case, we need to select n elements from the first part of the set. The number of ways to do this is given by the binomial coefficient nCn = 1.

Case 3: Choosing k elements from the first part, where k > n
In this case, we need to select k elements from the first part of the set, where k > n. The number of ways to do this is given by the sum of all binomial coefficients from n+1 to 2n+1, inclusive.
Sum = (n+1)C(n+1) + (n+1)C(n+2) + (n+1)C(n+3) + ... + (n+1)C(2n+1)

Using the identity (r-1)Cr + rCr = (r+1)Cr+1, we can simplify the above sum as follows:
Sum = (n+2)C(n+1) + (n+3)C(n+1) + ... + (2n+1)C(n+1)
= (2n+1+1)C(n+1) - (n+1)C(n+1)
= (2n+2)C(n+1) - 1C1
= (2n+2)C(n+1) - 1

Therefore, the total number of subsets that contain more than n elements is given by the sum of the cases:
Total = Case 1 + Case 2 + Case 3
= 1 + 1 + ((2n+2)C(n+1) - 1)
= (2n+2)C(n+1)

Now, we know that the number of subsets of a set with 2n+1 elements is 2^(2n+1). Therefore, the number of subsets that contain more than n elements is given by:
Total = (2