Evaluate :
(i) `""^(20)P_(4)`
(ii) `""^(75)P_(2)`

Answer

To evaluate the permutations given in the question, we will use the formula for permutations, which is defined as:

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

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

### Part (i): Evaluate \( 20P4 \)

1. **Identify \( n \) and \( r \)**:
- Here, \( n = 20 \) and \( r = 4 \).

2. **Apply the permutation formula**:
\[ 20P4 = \frac{20!}{(20-4)!} = \frac{20!}{16!} \]

3. **Simplify the factorial expression**:
- We can expand \( 20! \) as follows:
\[ 20! = 20 \times 19 \times 18 \times 17 \times 16! \]
- Therefore, substituting back, we get:
\[ 20P4 = \frac{20 \times 19 \times 18 \times 17 \times 16!}{16!} \]

4. **Cancel \( 16! \)**:
- The \( 16! \) in the numerator and denominator cancels out:
\[ 20P4 = 20 \times 19 \times 18 \times 17 \]

5. **Calculate the product**:
- Now we compute:
\[ 20 \times 19 = 380 \]
\[ 380 \times 18 = 6840 \]
\[ 6840 \times 17 = 116280 \]

Thus, \( 20P4 = 116280 \).

### Part (ii): Evaluate \( 75P2 \)

1. **Identify \( n \) and \( r \)**:
- Here, \( n = 75 \) and \( r = 2 \).

2. **Apply the permutation formula**:
\[ 75P2 = \frac{75!}{(75-2)!} = \frac{75!}{73!} \]

3. **Simplify the factorial expression**:
- We can expand \( 75! \) as follows:
\[ 75! = 75 \times 74 \times 73! \]
- Therefore, substituting back, we get:
\[ 75P2 = \frac{75 \times 74 \times 73!}{73!} \]

4. **Cancel \( 73! \)**:
- The \( 73! \) in the numerator and denominator cancels out:
\[ 75P2 = 75 \times 74 \]

5. **Calculate the product**:
- Now we compute:
\[ 75 \times 74 = 5550 \]

Thus, \( 75P2 = 5550 \).

### Final Answers:
- (i) \( 20P4 = 116280 \)
- (ii) \( 75P2 = 5550 \)