There are 15 couples taking part in a single tournament , the number of ways in which they can be paired such that no two real lives couple play in the same team is

Options

A. `15!(1-(1)/(1!)+(1)/(2!)-(1)/(3!)* * * * * * -(1)/(15!))`
B. `15!`
C. `""^(30)C_(2)`
D. `""^(30)C_(2)-""^(15)C_(1)`

Answer

To solve the problem of pairing 15 couples in such a way that no two real-life couples are on the same team, we can use the concept of derangements. A derangement is a permutation of elements such that none of the elements appear in their original position.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have 15 couples, which means there are 30 individuals. We need to pair them into teams of two such that no couple is paired together.

2. **Using the Derangement Formula**:
The number of ways to arrange \( n \) items such that no item appears in its original position is given by the formula for derangements, denoted as \( !n \):
\[ !n = n! \left( \sum_{i=0}^{n} \frac{(-1)^i}{i!} \right) \]
For our case, \( n = 15 \).

3. **Calculating \( !15 \)**:
We will substitute \( n = 15 \) into the derangement formula:
\[ !15 = 15! \left( \sum_{i=0}^{15} \frac{(-1)^i}{i!} \right) \]

4. **Calculating the Summation**:
We need to calculate the summation:
\[ \sum_{i=0}^{15} \frac{(-1)^i}{i!} = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} - \ldots + \frac{(-1)^{15}}{15!} \]
This series converges to \( e^{-1} \) as \( n \) approaches infinity, but we will compute it explicitly for \( n = 15 \).

5. **Final Calculation**:
After calculating the summation, we multiply it by \( 15! \) to find the total number of ways to pair the couples:
\[ !15 = 15! \left( \text{calculated sum} \right) \]

6. **Conclusion**:
The final result will give us the number of ways to pair the 15 couples such that no two real-life couples are on the same team.