Evaluate the following :
(i) `""^(8)P_(5)`
(ii) `""^(10)P_(3)`
(iii) `""^(12)C_(7)`
(iv) `""^(9)C_(0)`.

Answer

To solve the given problems step by step, we will use the formulas for permutations and combinations.

### Step-by-Step Solution:

**(i) Evaluate \( ^8P_5 \)**

The formula for permutations is given by:
\[ ^nP_r = \frac{n!}{(n-r)!} \]
For \( ^8P_5 \):
\[ ^8P_5 = \frac{8!}{(8-5)!} = \frac{8!}{3!} \]
Now, we calculate \( 8! \) and \( 3! \):
\[ 8! = 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 40320 \]
\[ 3! = 3 \times 2 \times 1 = 6 \]
Now substituting back:
\[ ^8P_5 = \frac{40320}{6} = 6720 \]

**(ii) Evaluate \( ^{10}P_3 \)**

Using the same formula:
\[ ^{10}P_3 = \frac{10!}{(10-3)!} = \frac{10!}{7!} \]
Calculating \( 10! \):
\[ 10! = 10 \times 9 \times 8 \times 7! \]
So,
\[ ^{10}P_3 = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8 = 720 \]

**(iii) Evaluate \( ^{12}C_7 \)**

The formula for combinations is given by:
\[ ^nC_r = \frac{n!}{r!(n-r)!} \]
For \( ^{12}C_7 \):
\[ ^{12}C_7 = \frac{12!}{7! \times 5!} \]
Calculating \( 12! \):
\[ 12! = 12 \times 11 \times 10 \times 9 \times 8 \times 7! \]
Now substituting:
\[ ^{12}C_7 = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7!}{7! \times 5!} \]
Cancelling \( 7! \):
\[ ^{12}C_7 = \frac{12 \times 11 \times 10 \times 9 \times 8}{5!} \]
Calculating \( 5! \):
\[ 5! = 120 \]
Now substituting back:
\[ ^{12}C_7 = \frac{12 \times 11 \times 10 \times 9 \times 8}{120} \]
Calculating the numerator:
\[ 12 \times 11 = 132 \]
\[ 132 \times 10 = 1320 \]
\[ 1320 \times 9 = 11880 \]
\[ 11880 \times 8 = 95040 \]
Now dividing by \( 120 \):
\[ ^{12}C_7 = \frac{95040}{120} = 792 \]

**(iv) Evaluate \( ^9C_0 \)**

Using the combination formula:
\[ ^nC_0 = \frac{n!}{0!(n-0)!} = \frac{n!}{0! \times n!} \]
Since \( 0! = 1 \):
\[ ^9C_0 = \frac{9!}{1 \times 9!} = 1 \]

### Final Answers:
1. \( ^8P_5 = 6720 \)
2. \( ^{10}P_3 = 720 \)
3. \( ^{12}C_7 = 792 \)
4. \( ^9C_0 = 1 \)