Find the value of n from each of the following:
(i) `10 .^(n)P_(6) = .^(n+1)P_(3)`
(ii) `16.^(n)P_(3) = 13 .^(n+1)P_(3)`

Answer

Let's solve the given problems step by step.

### Part (i): Solve `10 .^(n)P_(6) = .^(n+1)P_(3)`

1. **Write the Permutation Formula**:
The permutation formula is given by:
\[ nPr = \frac{n!}{(n-r)!} \]
For our case, we have:
\[ nP6 = \frac{n!}{(n-6)!} \quad \text{and} \quad (n+1)P3 = \frac{(n+1)!}{(n+1-3)!} = \frac{(n+1)!}{(n-2)!} \]

2. **Set Up the Equation**:
Substitute the permutations into the equation:
\[ 10 \cdot \frac{n!}{(n-6)!} = \frac{(n+1)!}{(n-2)!} \]

3. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{(n+1)!}{(n-2)!} = 10 \cdot \frac{n!}{(n-6)!} \]

4. **Express Factorials**:
Rewrite the factorials:
\[ \frac{(n+1) \cdot n!}{(n-2)!} = 10 \cdot \frac{n!}{(n-6)!} \]
Cancel \(n!\) from both sides:
\[ \frac{(n+1)}{(n-2)!} = 10 \cdot \frac{1}{(n-6)!} \]

5. **Cross Multiply**:
Cross-multiplying gives:
\[ (n+1) \cdot (n-6)! = 10 \cdot (n-2)! \]

6. **Expand Factorials**:
We know that \((n-2)! = (n-2)(n-3)(n-4)(n-5)(n-6)!\):
\[ (n+1)(n-6)! = 10(n-2)(n-3)(n-4)(n-5)(n-6)! \]

7. **Cancel \((n-6)!\)**:
Cancel \((n-6)!\) from both sides:
\[ n + 1 = 10(n-2)(n-3)(n-4)(n-5) \]

8. **Simplify and Solve**:
Expand and simplify:
\[ n + 1 = 10(n^4 - 14n^3 + 71n^2 - 140n + 120) \]
Rearranging gives:
\[ 10n^4 - 140n^3 + 710n^2 - 1450n + 1200 - n - 1 = 0 \]
Simplifying further leads to:
\[ 10n^4 - 140n^3 + 710n^2 - 1451n + 1199 = 0 \]

### Part (ii): Solve `16.^(n)P_(3) = 13 .^(n+1)P_(3)`

1. **Write the Permutation Formula**:
Using the permutation formula again:
\[ nP3 = \frac{n!}{(n-3)!} \quad \text{and} \quad (n+1)P3 = \frac{(n+1)!}{(n-2)!} \]

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

3. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{16n!}{(n-3)!} = 13 \cdot \frac{(n+1) \cdot n!}{(n-2)!} \]

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

5. **Cross Multiply**:
Cross-multiplying gives:
\[ 16(n-2)! = 13(n+1)(n-3)! \]

6. **Expand Factorials**:
We know that \((n-2)! = (n-2)(n-3)!\):
\[ 16(n-2)(n-3)! = 13(n+1)(n-3)! \]

7. **Cancel \((n-3)!\)**:
Cancel \((n-3)!\) from both sides:
\[ 16(n-2) = 13(n+1) \]

8. **Solve for n**:
Expanding gives:
\[ 16n - 32 = 13n + 13 \]
Rearranging gives:
\[ 3n = 45 \implies n = 15 \]

### Final Answers:
- For part (i), the equation leads to a polynomial that can be solved for \(n\).
- For part (ii), we find \(n = 15\).