(i) If ` ""^(n)P_(5)=20xx""^(n)P_(3)`, find n.
(ii) If `16xx""^(n)P_(3)=13xx""^(n+1)P_(3)`, find n.
(iii) If `""^(2n)P_(3)=100xx""^(n)P_(2)`, find n.

Answer

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

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

### (i) If \( nP_5 = 20 \times nP_3 \), find \( n \).

1. **Write the permutation formulas:**
\[ nP_5 = \frac{n!}{(n-5)!} \quad \text{and} \quad nP_3 = \frac{n!}{(n-3)!} \]

2. **Set up the equation:**
\[ \frac{n!}{(n-5)!} = 20 \times \frac{n!}{(n-3)!} \]

3. **Cancel \( n! \) from both sides:**
\[ \frac{1}{(n-5)!} = 20 \times \frac{1}{(n-3)!} \]

4. **Rewrite \( (n-3)! \):**
\[ (n-3)! = (n-3)(n-4)(n-5)! \]

5. **Substitute into the equation:**
\[ \frac{1}{(n-5)!} = \frac{20}{(n-3)(n-4)(n-5)!} \]

6. **Cross-multiply:**
\[ (n-3)(n-4) = 20 \]

7. **Expand and rearrange:**
\[ n^2 - 7n + 12 - 20 = 0 \implies n^2 - 7n - 8 = 0 \]

8. **Factor the quadratic:**
\[ (n - 8)(n + 1) = 0 \]

9. **Solve for \( n \):**
\[ n = 8 \quad \text{(we discard } n = -1 \text{ since } n \text{ must be positive)} \]

### (ii) If \( 16 \times nP_3 = 13 \times (n+1)P_3 \), find \( n \).

1. **Write the permutation formulas:**
\[ nP_3 = \frac{n!}{(n-3)!} \quad \text{and} \quad (n+1)P_3 = \frac{(n+1)!}{(n-2)!} \]

2. **Set up the equation:**
\[ 16 \times \frac{n!}{(n-3)!} = 13 \times \frac{(n+1)!}{(n-2)!} \]

3. **Rewrite \( (n+1)! \):**
\[ (n+1)! = (n+1) \times n! \]

4. **Substitute into the equation:**
\[ 16 \times \frac{n!}{(n-3)!} = 13 \times \frac{(n+1) \times n!}{(n-2)!} \]

5. **Cancel \( n! \) from both sides:**
\[ 16 \times \frac{1}{(n-3)!} = 13 \times \frac{(n+1)}{(n-2)(n-3)!} \]

6. **Cross-multiply:**
\[ 16(n-2) = 13(n+1) \]

7. **Expand and rearrange:**
\[ 16n - 32 = 13n + 13 \implies 3n = 45 \implies n = 15 \]

### (iii) If \( 2nP_3 = 100 \times nP_2 \), find \( n \).

1. **Write the permutation formulas:**
\[ 2nP_3 = \frac{(2n)!}{(2n-3)!} \quad \text{and} \quad nP_2 = \frac{n!}{(n-2)!} \]

2. **Set up the equation:**
\[ \frac{(2n)!}{(2n-3)!} = 100 \times \frac{n!}{(n-2)!} \]

3. **Rewrite \( (2n)! \):**
\[ (2n)! = (2n)(2n-1)(2n-2)(2n-3)! \]

4. **Substitute into the equation:**
\[ (2n)(2n-1)(2n-2) = 100 \times \frac{n!}{(n-2)!} \]

5. **Rewrite \( n! \):**
\[ n! = n(n-1)(n-2)! \]

6. **Substitute into the equation:**
\[ (2n)(2n-1)(2n-2) = 100 \times n(n-1) \]

7. **Expand and rearrange:**
\[ 8n^3 - 12n^2 + 4n = 100n^2 - 100n \]

8. **Combine like terms:**
\[ 8n^3 - 112n^2 + 104n = 0 \]

9. **Factor out \( n \):**
\[ n(8n^2 - 112n + 104) = 0 \]

10. **Solve the quadratic:**
\[ 8n^2 - 112n + 104 = 0 \implies n^2 - 14n + 13 = 0 \]

11. **Factor the quadratic:**
\[ (n - 13)(n - 1) = 0 \]

12. **Solve for \( n \):**
\[ n = 13 \quad \text{or} \quad n = 1 \]

### Summary of Solutions:
- (i) \( n = 8 \)
- (ii) \( n = 15 \)
- (iii) \( n = 13 \) or \( n = 1 \)