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 S1 and S2 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 up 16 players into 8 pairs can be calculated using the formula for pairing \( n \) items:
\[ \text{Total ways} = \frac{n!}{2^k \cdot k!} \]
where \( n \) is the total number of players and \( k \) is the number of pairs. Here, \( n = 16 \) and \( k = 8 \).
\[ \text{Total ways} = \frac{16!}{2^8 \cdot 8!} \]

### Step 2: Calculate the number of ways in which S1 and S2 are in the same group.
If S1 and S2 are in the same group, we can treat them as a single entity or a "super player." This reduces the problem to pairing 15 entities (S1 and S2 as one entity and the other 14 players).
\[ \text{Ways with S1 and S2 together} = \frac{14!}{2^7 \cdot 7!} \]

### Step 3: Calculate the required number of ways.
To find the number of ways in which S1 and S2 are in different groups, we subtract the number of ways they are in the same group from the total number of ways:
\[ \text{Required ways} = \text{Total ways} - \text{Ways with S1 and S2 together} \]
Substituting the values we found:
\[ \text{Required ways} = \frac{16!}{2^8 \cdot 8!} - \frac{14!}{2^7 \cdot 7!} \]

### Step 4: Simplify the expression.
To simplify the expression:
1. Factor out \( \frac{14!}{2^7 \cdot 7!} \) from both terms:
\[ \text{Required ways} = \frac{14!}{2^7 \cdot 7!} \left( \frac{16 \cdot 15}{2 \cdot 8} - 1 \right) \]
2. Simplify the term inside the parentheses:
\[ \frac{16 \cdot 15}{2 \cdot 8} = \frac{240}{16} = 15 \]
Thus:
\[ \text{Required ways} = \frac{14!}{2^7 \cdot 7!} (15 - 1) = \frac{14!}{2^7 \cdot 7!} \cdot 14 \]

### Step 5: Final expression.
The final expression for the number of ways to group the players such that S1 and S2 are in different groups is:
\[ \text{Required ways} = \frac{14! \cdot 14}{2^7 \cdot 7!} \]