Find n if :
` "" ^(2n)P_(n+1) : ""^(2n-2) P_n = 56 : 3 `

Answer

To solve the problem, we need to find the value of \( n \) given the equation:

\[ \frac{(2n)P_{n+1}}{(2n-2)P_n} = \frac{56}{3} \]

### Step-by-step Solution:

1. **Write the Permutation Formula**:
The permutation \( nPk \) is given by the formula:
\[ nPk = \frac{n!}{(n-k)!} \]
Therefore, we can express \( (2n)P_{n+1} \) and \( (2n-2)P_n \) using this formula.

2. **Express the Permutations**:
\[ (2n)P_{n+1} = \frac{(2n)!}{(2n-(n+1))!} = \frac{(2n)!}{(n-1)!} \]
\[ (2n-2)P_n = \frac{(2n-2)!}{(2n-2-n)!} = \frac{(2n-2)!}{(n-2)!} \]

3. **Substituting into the Equation**:
Substitute these expressions back into the original equation:
\[ \frac{\frac{(2n)!}{(n-1)!}}{\frac{(2n-2)!}{(n-2)!}} = \frac{56}{3} \]

4. **Simplifying the Left Side**:
This simplifies to:
\[ \frac{(2n)! \cdot (n-2)!}{(n-1)! \cdot (2n-2)!} \]
Now, we can use the identity \( (2n)! = (2n)(2n-1)(2n-2)! \):
\[ \frac{(2n)(2n-1)(2n-2)! \cdot (n-2)!}{(n-1)! \cdot (2n-2)!} = \frac{(2n)(2n-1)(n-2)!}{(n-1)!} \]

5. **Cancelling Factorials**:
The \( (2n-2)! \) cancels out:
\[ \frac{(2n)(2n-1)}{(n-1)} = \frac{56}{3} \]

6. **Cross-Multiplying**:
Cross-multiply to eliminate the fraction:
\[ 3(2n)(2n-1) = 56(n-1) \]
Expanding both sides gives:
\[ 12n^2 - 6n = 56n - 56 \]

7. **Rearranging the Equation**:
Bring all terms to one side:
\[ 12n^2 - 62n + 56 = 0 \]

8. **Dividing the Equation**:
Divide the entire equation by 2 for simplification:
\[ 6n^2 - 31n + 28 = 0 \]

9. **Factoring the Quadratic**:
We need to factor this quadratic. We look for two numbers that multiply to \( 6 \times 28 = 168 \) and add to \( -31 \). The numbers are \( -24 \) and \( -7 \):
\[ 6n^2 - 24n - 7n + 28 = 0 \]
Grouping gives:
\[ (6n - 7)(n - 4) = 0 \]

10. **Finding the Roots**:
Setting each factor to zero gives:
\[ 6n - 7 = 0 \quad \Rightarrow \quad n = \frac{7}{6} \quad (\text{not valid since } n \text{ must be an integer}) \]
\[ n - 4 = 0 \quad \Rightarrow \quad n = 4 \]

### Final Answer:
Thus, the value of \( n \) is:
\[ \boxed{4} \]