If `.^(n+2)C_(8) : ^(n-2)P_(4)=57:16`, then the value of `(n)/(2)` is

Answer

To solve the problem, we need to find the value of \( \frac{n}{2} \) given the ratio \( \binom{n+2}{8} : P(n-2, 4) = 57 : 16 \).

### Step-by-Step Solution:

1. **Understanding the Combinations and Permutations**:
- The combination \( \binom{n+2}{8} \) is given by the formula:
\[ \binom{n+2}{8} = \frac{(n+2)!}{8!(n-6)!} \]
- The permutation \( P(n-2, 4) \) is given by the formula:
\[ P(n-2, 4) = \frac{(n-2)!}{(n-6)!} \]

2. **Setting Up the Ratio**:
- We can express the ratio as:
\[ \frac{\binom{n+2}{8}}{P(n-2, 4)} = \frac{\frac{(n+2)!}{8!(n-6)!}}{\frac{(n-2)!}{(n-6)!}} = \frac{(n+2)!}{8!(n-2)!} \]
- Thus, we have:
\[ \frac{(n+2)!}{8!(n-2)!} = \frac{57}{16} \]

3. **Cross-Multiplying**:
- Cross-multiplying gives us:
\[ 16(n+2)! = 57 \cdot 8!(n-2)! \]

4. **Expanding Factorials**:
- We can express \( (n+2)! \) as:
\[ (n+2)(n+1)(n)! \]
- And \( (n-2)! \) can be expressed as:
\[ (n-2)! = \frac{(n)!}{(n)(n-1)} \]
- Substituting these into our equation gives:
\[ 16(n+2)(n+1)(n)! = 57 \cdot 8! \cdot \frac{(n)!}{(n)(n-1)} \]

5. **Cancelling \( n! \)**:
- Since \( n! \) is common on both sides, we can cancel it out (assuming \( n \neq 0 \)):
\[ 16(n+2)(n+1) = \frac{57 \cdot 8!}{(n)(n-1)} \]

6. **Calculating \( 8! \)**:
- We know \( 8! = 40320 \), so substituting this gives:
\[ 16(n+2)(n+1) = \frac{57 \cdot 40320}{(n)(n-1)} \]

7. **Simplifying**:
- Rearranging gives:
\[ 16(n+2)(n+1)(n)(n-1) = 57 \cdot 40320 \]

8. **Finding \( n \)**:
- This is a polynomial equation in \( n \). Solving this will yield the value of \( n \). After solving, we find that \( n = 19 \).

9. **Finding \( \frac{n}{2} \)**:
- Finally, we calculate:
\[ \frac{n}{2} = \frac{19}{2} = 9.5 \]

### Final Answer:
\[ \frac{n}{2} = 9.5 \]