If `.^(n)P_(5)=20 .^(n)P_(3)`, find the value of n.

Explanation

To solve the equation \( nP_5 = 20 \cdot nP_3 \), we will follow these steps:

### Step 1: Write the formula for permutations
The permutation formula is given by:
\[ nP_r = \frac{n!}{(n-r)!} \]
Using this, we can express \( nP_5 \) and \( nP_3 \).

### Step 2: Substitute the permutation values
Substituting into the equation:
\[ nP_5 = \frac{n!}{(n-5)!} \]
\[ nP_3 = \frac{n!}{(n-3)!} \]
Thus, the equation becomes:
\[ \frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!} \]

### Step 3: Simplify the equation
We can cancel \( n! \) from both sides (assuming \( n! \neq 0 \)):
\[ \frac{1}{(n-5)!} = 20 \cdot \frac{1}{(n-3)!} \]
This simplifies to:
\[ (n-3)! = 20 \cdot (n-5)! \]

### Step 4: Express \( (n-3)! \) in terms of \( (n-5)! \)
We can express \( (n-3)! \) as:
\[ (n-3)! = (n-3)(n-4)(n-5)! \]
Substituting this into the equation gives:
\[ (n-3)(n-4)(n-5)! = 20 \cdot (n-5)! \]

### Step 5: Cancel \( (n-5)! \)
Assuming \( (n-5)! \neq 0 \), we can cancel \( (n-5)! \) from both sides:
\[ (n-3)(n-4) = 20 \]

### Step 6: Expand and rearrange the equation
Expanding the left side:
\[ n^2 - 7n + 12 = 20 \]
Rearranging gives:
\[ n^2 - 7n - 8 = 0 \]

### Step 7: Factor the quadratic equation
We can factor the quadratic equation:
\[ (n - 8)(n + 1) = 0 \]

### Step 8: Solve for \( n \)
Setting each factor to zero gives:
\[ n - 8 = 0 \quad \Rightarrow \quad n = 8 \]
\[ n + 1 = 0 \quad \Rightarrow \quad n = -1 \]
Since \( n \) must be a non-negative integer, we discard \( n = -1 \).

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{8} \]