Evaluate the following:
(i) `.^(9)P_(3)` (ii) `.^(10)P_(2)`
(iii) `.^(12)P_(4)`

Answer

To evaluate the permutations given in the question, we will use the formula for permutations:

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

where \( n \) is the total number of items, \( r \) is the number of items to arrange, and \( ! \) denotes factorial.

### Step-by-Step Solution:

**(i) Evaluate \( ^9P_3 \)**

1. **Identify \( n \) and \( r \)**: Here, \( n = 9 \) and \( r = 3 \).
2. **Apply the permutation formula**:
\[ ^9P_3 = \frac{9!}{(9-3)!} = \frac{9!}{6!} \]
3. **Expand \( 9! \)**:
\[ 9! = 9 \times 8 \times 7 \times 6! \]
4. **Substitute back into the equation**:
\[ ^9P_3 = \frac{9 \times 8 \times 7 \times 6!}{6!} \]
5. **Cancel \( 6! \)**:
\[ ^9P_3 = 9 \times 8 \times 7 \]
6. **Calculate the product**:
\[ 9 \times 8 = 72 \]
\[ 72 \times 7 = 504 \]
Thus, \( ^9P_3 = 504 \).

**(ii) Evaluate \( ^{10}P_2 \)**

1. **Identify \( n \) and \( r \)**: Here, \( n = 10 \) and \( r = 2 \).
2. **Apply the permutation formula**:
\[ ^{10}P_2 = \frac{10!}{(10-2)!} = \frac{10!}{8!} \]
3. **Expand \( 10! \)**:
\[ 10! = 10 \times 9 \times 8! \]
4. **Substitute back into the equation**:
\[ ^{10}P_2 = \frac{10 \times 9 \times 8!}{8!} \]
5. **Cancel \( 8! \)**:
\[ ^{10}P_2 = 10 \times 9 \]
6. **Calculate the product**:
\[ 10 \times 9 = 90 \]
Thus, \( ^{10}P_2 = 90 \).

**(iii) Evaluate \( ^{12}P_4 \)**

1. **Identify \( n \) and \( r \)**: Here, \( n = 12 \) and \( r = 4 \).
2. **Apply the permutation formula**:
\[ ^{12}P_4 = \frac{12!}{(12-4)!} = \frac{12!}{8!} \]
3. **Expand \( 12! \)**:
\[ 12! = 12 \times 11 \times 10 \times 9 \times 8! \]
4. **Substitute back into the equation**:
\[ ^{12}P_4 = \frac{12 \times 11 \times 10 \times 9 \times 8!}{8!} \]
5. **Cancel \( 8! \)**:
\[ ^{12}P_4 = 12 \times 11 \times 10 \times 9 \]
6. **Calculate the product**:
\[ 12 \times 11 = 132 \]
\[ 132 \times 10 = 1320 \]
\[ 1320 \times 9 = 11880 \]
Thus, \( ^{12}P_4 = 11880 \).

### Final Answers:
- \( ^9P_3 = 504 \)
- \( ^{10}P_2 = 90 \)
- \( ^{12}P_4 = 11880 \)