Find 'r' if :
(i) `""^(5)P_(r )= ""^(6)P_(r-1)`
(ii) `5""^(4)P_(r )=6""^(5)P_(r-1)`
(iii) `""^(5)P_(r )=2""^(6)P_(r-1)`
(iv) `""^(10)P_(r+1): ""^(11)P_(r )=30 :11`.

Answer

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

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

### (i) Find \( r \) if \( ^5P_r = ^6P_{r-1} \)

1. **Write the permutation formulas:**
\[ ^5P_r = \frac{5!}{(5-r)!} \]
\[ ^6P_{r-1} = \frac{6!}{(6-(r-1))!} = \frac{6!}{(7-r)!} \]

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

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

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

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 simplify:**
\[ 42 - 13r + r^2 = 6 \]
\[ r^2 - 13r + 36 = 0 \]

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

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

11. **Check validity:**
Since \( r \) must be less than or equal to 5, the only valid solution is:
\[ r = 4 \]