Let `S={1,2,3, . . .,n}`. If X denotes the set of all subsets of S containing exactly two elements, then the value of `sum_(A in X) ` (min. A) is given by

Options

A. `.^(n+1)C_(3)`
B. `.^(n)C_(3)`
C. `(n(n^(2)-1))/(6)`
D. `(n(n^(2)-3n+2))/(6)`

Explanation

To solve the problem, we need to find the value of the summation of the minimum element of all subsets of the set \( S = \{1, 2, 3, \ldots, n\} \) that contain exactly two elements. Let's denote the set of all such subsets as \( X \).

### Step-by-Step Solution:

1. **Understanding the Set \( X \)**:
The set \( X \) consists of all subsets of \( S \) that contain exactly two elements. For example, if \( n = 3 \), then \( X = \{\{1, 2\}, \{1, 3\}, \{2, 3\}\} \).

2. **Identifying the Minimum Element**:
For each subset \( A = \{a, b\} \) in \( X \) where \( a < b \), the minimum element is \( a \). We need to sum \( \min(A) \) for all subsets \( A \) in \( X \).

3. **Counting Contributions of Each Element**:
Each element \( r \) in the set \( S \) can be the minimum of the subset \( A \). If \( r \) is the minimum, the other element \( b \) can be any element from the set \( \{r+1, r+2, \ldots, n\} \). The number of choices for \( b \) is \( n - r \).

4. **Setting Up the Summation**:
Therefore, the contribution of each element \( r \) to the sum is \( r \times (n - r) \). We can express the total sum as:
\[ \sum_{A \in X} \min(A) = \sum_{r=1}^{n-1} r(n - r) \]

5. **Expanding the Summation**:
We can expand the summation:
\[ \sum_{r=1}^{n-1} r(n - r) = \sum_{r=1}^{n-1} (nr - r^2) \]
This can be separated into two sums:
\[ = n \sum_{r=1}^{n-1} r - \sum_{r=1}^{n-1} r^2 \]

6. **Calculating the Sums**:
We use the formulas for the sums of the first \( m \) natural numbers and the sum of the squares:
- The sum of the first \( m \) natural numbers:
\[ \sum_{r=1}^{m} r = \frac{m(m+1)}{2} \]
- The sum of the squares of the first \( m \) natural numbers:
\[ \sum_{r=1}^{m} r^2 = \frac{m(m+1)(2m+1)}{6} \]
For \( m = n-1 \):
- \( \sum_{r=1}^{n-1} r = \frac{(n-1)n}{2} \)
- \( \sum_{r=1}^{n-1} r^2 = \frac{(n-1)n(2(n-1)+1)}{6} = \frac{(n-1)n(2n-1)}{6} \)

7. **Substituting Back**:
Substitute these results back into our expression:
\[ = n \cdot \frac{(n-1)n}{2} - \frac{(n-1)n(2n-1)}{6} \]

8. **Finding a Common Denominator**:
The common denominator for the two terms is 6:
\[ = \frac{3n(n-1)n}{6} - \frac{(n-1)n(2n-1)}{6} \]
\[ = \frac{(n-1)n(3n - (2n - 1))}{6} \]
\[ = \frac{(n-1)n(n + 1)}{6} \]

9. **Final Result**:
Thus, the value of the summation \( \sum_{A \in X} \min(A) \) is:
\[ \frac{(n-1)n(n + 1)}{6} \]