In how many ways can 11 players be selected from 14 players if
(i) a particular player is always included?
(ii) a particular player is never included?

Answer

To solve the problem of selecting 11 players from 14 players under the given conditions, we will break it down into two parts.

### Part (i): A particular player is always included

1. **Understanding the problem**: Since one particular player is always included, we only need to select 10 more players from the remaining 13 players.

2. **Choosing players**: The number of ways to choose 10 players from 13 can be represented using combinations, denoted as \( \binom{n}{r} \), where \( n \) is the total number of players to choose from, and \( r \) is the number of players to choose.

\[ \text{Number of ways} = \binom{13}{10} \]

3. **Using the combination formula**: The formula for combinations is given by:

\[ \binom{n}{r} = \frac{n!}{r!(n-r)!} \]

Applying this to our case:

\[ \binom{13}{10} = \frac{13!}{10! \cdot (13-10)!} = \frac{13!}{10! \cdot 3!} \]

4. **Calculating the factorials**: We can simplify this expression:

\[ \binom{13}{10} = \frac{13 \times 12 \times 11 \times 10!}{10! \times 3 \times 2 \times 1} \]

The \( 10! \) cancels out:

\[ = \frac{13 \times 12 \times 11}{3 \times 2 \times 1} \]

5. **Performing the calculations**:

\[ = \frac{13 \times 12 \times 11}{6} = \frac{1716}{6} = 286 \]

Thus, the number of ways to select 11 players when a particular player is always included is **286**.

### Part (ii): A particular player is never included

1. **Understanding the problem**: If a particular player is never included, we need to select all 11 players from the remaining 13 players.

2. **Choosing players**: The number of ways to choose 11 players from 13 is given by:

\[ \text{Number of ways} = \binom{13}{11} \]

3. **Using the combination formula**: Using the same combination formula:

\[ \binom{13}{11} = \frac{13!}{11! \cdot (13-11)!} = \frac{13!}{11! \cdot 2!} \]

4. **Calculating the factorials**: We can simplify this expression:

\[ \binom{13}{11} = \frac{13 \times 12 \times 11!}{11! \times 2!} \]

The \( 11! \) cancels out:

\[ = \frac{13 \times 12}{2 \times 1} \]

5. **Performing the calculations**:

\[ = \frac{156}{2} = 78 \]

Thus, the number of ways to select 11 players when a particular player is never included is **78**.

### Final Answers:
- (i) 286
- (ii) 78