In how many ways can 2n students be divided into n pairs?

Options

A. `n^2`
B. `(n!)/(2)`
C. `(2n!)/((2!)^n .(n!))`
D. none of these

Answer

To find the number of ways to divide 2n students into n pairs, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have 2n students and we want to form n pairs. Each pair consists of 2 students.

2. **Arranging the Students**:
The total number of ways to arrange 2n students is given by \( (2n)! \). This is because we can arrange all students in any order.

3. **Forming Pairs**:
Once we have arranged the 2n students, we can form pairs from this arrangement. The first two students in the arrangement will form the first pair, the next two will form the second pair, and so on. Therefore, we will have n pairs.

4. **Accounting for Order in Pairs**:
However, within each pair, the order of the two students does not matter. For example, the pair (A, B) is the same as (B, A). Since there are 2 students in each pair, we need to divide by \( 2! \) for each of the n pairs to account for this. Thus, we divide by \( (2!)^n \).

5. **Accounting for the Order of Pairs**:
Additionally, the order of the pairs themselves does not matter. For example, the arrangement of pairs (A, B) and (C, D) is the same as (C, D) and (A, B). Since there are n pairs, we also need to divide by \( n! \).

6. **Final Formula**:
Combining all these factors, the total number of ways to divide 2n students into n pairs is given by:
\[ \text{Number of ways} = \frac{(2n)!}{(2!)^n \cdot n!} \]

### Conclusion:
Thus, the number of ways to divide 2n students into n pairs is:
\[ \frac{(2n)!}{(2^n) \cdot n!} \]