Find `r` if
(i) `""^5P_r=2^6P_(r-1)`
(ii) `""^5P_r=^6P_(r-1)`

Options

A. a. 3
b. 9
B. a. 10
b. 4
C. a. 3
b. 4
D. a. 4
b. 3

Answer

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

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

### Part (i): Solve \( ^5P_r = 2 \times ^6P_{r-1} \)

1. **Write the formula for \( ^5P_r \)**:
\[ ^5P_r = \frac{5!}{(5-r)!} \]

2. **Write the formula for \( ^6P_{r-1} \)**:
\[ ^6P_{r-1} = \frac{6!}{(6-(r-1))!} = \frac{6!}{(7-r)!} \]

3. **Substituting the formulas into the equation**:
\[ \frac{5!}{(5-r)!} = 2 \times \frac{6!}{(7-r)!} \]

4. **Simplify \( 6! \)**:
\[ 6! = 6 \times 5! \]
So, we can rewrite the equation as:
\[ \frac{5!}{(5-r)!} = 2 \times \frac{6 \times 5!}{(7-r)!} \]

5. **Cancel \( 5! \)** from both sides**:
\[ \frac{1}{(5-r)!} = \frac{12}{(7-r)!} \]

6. **Cross-multiply**:
\[ (7-r)! = 12 \times (5-r)! \]

7. **Expand \( (7-r)! \)**:
\[ (7-r)(6-r)(5-r)! = 12 \times (5-r)! \]

8. **Cancel \( (5-r)! \)**:
\[ (7-r)(6-r) = 12 \]

9. **Expand and rearrange**:
\[ 42 - 13r + r^2 = 12 \]
\[ r^2 - 13r + 30 = 0 \]

10. **Factor the quadratic equation**:
\[ (r - 10)(r - 3) = 0 \]

11. **Find the values of \( r \)**:
\[ r = 10 \quad \text{or} \quad r = 3 \]

12. **Since \( r \) must be less than or equal to 5**:
\[ r = 3 \quad \text{is the valid solution.} \]

### Part (ii): Solve \( ^5P_r = ^6P_{r-1} \)

1. **Using the previous definitions**:
\[ ^5P_r = \frac{5!}{(5-r)!} \]
\[ ^6P_{r-1} = \frac{6!}{(7-r)!} \]

2. **Set the equations equal**:
\[ \frac{5!}{(5-r)!} = \frac{6!}{(7-r)!} \]

3. **Substituting \( 6! \)**:
\[ \frac{5!}{(5-r)!} = \frac{6 \times 5!}{(7-r)!} \]

4. **Cancel \( 5! \)**:
\[ \frac{1}{(5-r)!} = \frac{6}{(7-r)!} \]

5. **Cross-multiply**:
\[ (7-r)! = 6 \times (5-r)! \]

6. **Expand \( (7-r)! \)**:
\[ (7-r)(6-r)(5-r)! = 6 \times (5-r)! \]

7. **Cancel \( (5-r)! \)**:
\[ (7-r)(6-r) = 6 \]

8. **Expand and rearrange**:
\[ 42 - 13r + r^2 = 6 \]
\[ r^2 - 13r + 36 = 0 \]

9. **Factor the quadratic equation**:
\[ (r - 9)(r - 4) = 0 \]

10. **Find the values of \( r \)**:
\[ r = 9 \quad \text{or} \quad r = 4 \]

11. **Since \( r \) must be less than or equal to 5**:
\[ r = 4 \quad \text{is the valid solution.} \]

### Final Answers:
- For part (i): \( r = 3 \)
- For part (ii): \( r = 4 \)