Sixteen players `S_(1)`, `S_(2)`, `S_(3)`,…,`S_(16)` play in a tournament. Number of ways in which they can be grouped into eight pairs so that `S_(1)` and `S_(2)` are in different groups, is equal to

Options

A. `((14)!)/(2^(6)*6!)`
B. `((15)!)/(2^(7)*7!)`
C. `((14)!)/(2^(7)*6!)`
D. `((14)!)/(2^(6)*7!)`

Explanation

To solve the problem of grouping 16 players into 8 pairs such that players \( S_1 \) and \( S_2 \) are in different groups, we can follow these steps:

### Step 1: Calculate the total number of ways to pair 16 players
The total number of ways to pair 16 players into 8 pairs can be calculated using the formula:

\[ \text{Total ways} = \frac{16!}{2^8 \times 8!} \]

Where:
- \( 16! \) is the total arrangements of 16 players.
- \( 2^8 \) accounts for the fact that within each pair, the order does not matter (i.e., \( (A, B) \) is the same as \( (B, A) \)).
- \( 8! \) accounts for the arrangement of the 8 pairs.

### Step 2: Calculate the number of ways when \( S_1 \) and \( S_2 \) are in the same group
If \( S_1 \) and \( S_2 \) are in the same group, we can treat them as a single unit or pair. This leaves us with 14 players (the pair \( (S_1, S_2) \) plus the remaining 14 players).

The number of ways to pair these 15 units (the pair \( (S_1, S_2) \) and the 14 other players) is given by:

\[ \text{Ways with } S_1 \text{ and } S_2 \text{ together} = \frac{14!}{2^7 \times 7!} \]

Where:
- \( 14! \) is the total arrangements of the 14 remaining players plus the pair.
- \( 2^7 \) accounts for the order within the 7 pairs.
- \( 7! \) accounts for the arrangement of the 7 pairs.

### Step 3: Subtract the two results
To find the number of ways in which \( S_1 \) and \( S_2 \) are in different groups, we subtract the number of ways they are together from the total number of ways:

\[ \text{Required ways} = \frac{16!}{2^8 \times 8!} - \frac{14!}{2^7 \times 7!} \]

### Step 4: Simplify the expression
Now we can simplify the expression:

1. Factor out \( \frac{14!}{2^7 \times 7!} \):

\[ \text{Required ways} = \frac{14!}{2^7 \times 7!} \left( \frac{16!}{14! \times 2 \times 8} - 1 \right) \]

2. Simplifying \( \frac{16!}{14!} = 16 \times 15 \):

\[ = \frac{14!}{2^7 \times 7!} \left( \frac{16 \times 15}{2 \times 8} - 1 \right) \]

3. Calculate \( \frac{16 \times 15}{16} = 15 \):

\[ = \frac{14!}{2^7 \times 7!} (15 - 1) = \frac{14!}{2^7 \times 7!} \times 14 \]

4. Finally, we can express it as:

\[ = \frac{14!}{2^6 \times 6!} \]

### Final Answer
Thus, the number of ways in which the players can be grouped into pairs such that \( S_1 \) and \( S_2 \) are in different groups is:

\[ \frac{14!}{2^6 \times 6!} \]