Evaluate:
` "(i) "^(10)P_(4) " (ii) "^(62)P_(3)" (iii) "^(6)P_(6) " (iv) "^(9)P_(0)`

Answer

To solve the given permutations, we will use the formula for permutations, which is:

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

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

Now, let's evaluate each part step by step.

### (i) Evaluate \( ^{10}P_4 \)

1. **Identify \( n \) and \( r \)**: Here, \( n = 10 \) and \( r = 4 \).
2. **Apply the formula**:
\[ ^{10}P_4 = \frac{10!}{(10-4)!} = \frac{10!}{6!} \]
3. **Expand \( 10! \)**:
\[ 10! = 10 \times 9 \times 8 \times 7 \times 6! \]
4. **Substitute back into the equation**:
\[ ^{10}P_4 = \frac{10 \times 9 \times 8 \times 7 \times 6!}{6!} \]
5. **Cancel \( 6! \)**:
\[ ^{10}P_4 = 10 \times 9 \times 8 \times 7 \]
6. **Calculate the product**:
\[ 10 \times 9 = 90 \]
\[ 90 \times 8 = 720 \]
\[ 720 \times 7 = 5040 \]
7. **Final result**:
\[ ^{10}P_4 = 5040 \]

### (ii) Evaluate \( ^{62}P_3 \)

1. **Identify \( n \) and \( r \)**: Here, \( n = 62 \) and \( r = 3 \).
2. **Apply the formula**:
\[ ^{62}P_3 = \frac{62!}{(62-3)!} = \frac{62!}{59!} \]
3. **Expand \( 62! \)**:
\[ 62! = 62 \times 61 \times 60 \times 59! \]
4. **Substitute back into the equation**:
\[ ^{62}P_3 = \frac{62 \times 61 \times 60 \times 59!}{59!} \]
5. **Cancel \( 59! \)**:
\[ ^{62}P_3 = 62 \times 61 \times 60 \]
6. **Calculate the product**:
\[ 62 \times 61 = 3782 \]
\[ 3782 \times 60 = 226920 \]
7. **Final result**:
\[ ^{62}P_3 = 226920 \]

### (iii) Evaluate \( ^{6}P_6 \)

1. **Identify \( n \) and \( r \)**: Here, \( n = 6 \) and \( r = 6 \).
2. **Apply the formula**:
\[ ^{6}P_6 = \frac{6!}{(6-6)!} = \frac{6!}{0!} \]
3. **Recall that \( 0! = 1 \)**:
\[ ^{6}P_6 = 6! \]
4. **Calculate \( 6! \)**:
\[ 6! = 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \]
5. **Final result**:
\[ ^{6}P_6 = 720 \]

### (iv) Evaluate \( ^{9}P_0 \)

1. **Identify \( n \) and \( r \)**: Here, \( n = 9 \) and \( r = 0 \).
2. **Apply the formula**:
\[ ^{9}P_0 = \frac{9!}{(9-0)!} = \frac{9!}{9!} \]
3. **Simplify**:
\[ ^{9}P_0 = 1 \]
4. **Final result**:
\[ ^{9}P_0 = 1 \]

### Summary of Results:
- \( ^{10}P_4 = 5040 \)
- \( ^{62}P_3 = 226920 \)
- \( ^{6}P_6 = 720 \)
- \( ^{9}P_0 = 1 \)