Find thevalue of 'r':
(i) `.^(12)P_(r) = 1320`
(ii) `5.^(4)P_(r) = 6.^(5)P_(r-1)`

Answer

To solve the given problems, we will break them down step by step.

### (i) Find the value of 'r' for \( 12P_r = 1320 \)

1. **Understand the formula for permutations**:
The formula for permutations is given by:
\[ nP_r = \frac{n!}{(n-r)!} \]
For our case, \( n = 12 \) and we have:
\[ 12P_r = \frac{12!}{(12-r)!} \]
Setting this equal to 1320:
\[ \frac{12!}{(12-r)!} = 1320 \]

2. **Simplify the equation**:
We can rewrite the left-hand side:
\[ 12 \times 11 \times 10 \times \ldots \times (12 - r + 1) = 1320 \]
This means we need to find the product of \( r \) consecutive numbers starting from 12 downwards.

3. **Testing values for r**:
We will test integer values for \( r \) starting from 1 up to 12 to find when the product equals 1320.

- For \( r = 1 \):
\[ 12P_1 = 12 \quad (\text{not } 1320) \]
- For \( r = 2 \):
\[ 12P_2 = 12 \times 11 = 132 \quad (\text{not } 1320) \]
- For \( r = 3 \):
\[ 12P_3 = 12 \times 11 \times 10 = 1320 \quad (\text{this is correct!}) \]

4. **Conclusion for part (i)**:
The value of \( r \) is \( 3 \).

### (ii) Find the value of 'r' for \( 5P_r = 6P_{r-1} \)

1. **Write down the permutation formulas**:
Using the permutation formula:
\[ 5P_r = \frac{5!}{(5-r)!} \quad \text{and} \quad 6P_{r-1} = \frac{6!}{(6-(r-1))!} = \frac{6!}{(7-r)!} \]

2. **Set up the equation**:
We can set up the equation:
\[ \frac{5!}{(5-r)!} = \frac{6!}{(7-r)!} \]

3. **Cross-multiply**:
This gives us:
\[ 5! \cdot (7-r)! = 6! \cdot (5-r)! \]

4. **Substituting factorials**:
We know \( 6! = 6 \times 5! \), so substituting we get:
\[ 5! \cdot (7-r)! = 6 \cdot 5! \cdot (5-r)! \]
Dividing both sides by \( 5! \) (assuming \( 5! \neq 0 \)):
\[ (7-r)! = 6 \cdot (5-r)! \]

5. **Expanding the factorials**:
We can write:
\[ (7-r)(6-r)(5-r)! = 6 \cdot (5-r)! \]
Dividing both sides by \( (5-r)! \):
\[ (7-r)(6-r) = 6 \]

6. **Expanding and rearranging**:
Expanding gives:
\[ 42 - 13r + r^2 = 6 \]
Rearranging gives:
\[ r^2 - 13r + 36 = 0 \]

7. **Factoring the quadratic**:
We need two numbers that multiply to 36 and add to -13. The factors are:
\[ (r - 9)(r - 4) = 0 \]
Thus, \( r = 9 \) or \( r = 4 \).

8. **Conclusion for part (ii)**:
The values of \( r \) are \( 4 \) and \( 9 \).

### Final Answers:
(i) \( r = 3 \)
(ii) \( r = 4 \) or \( r = 9 \)