find the value of n, if ` ""^(n) P_5 = 20 ""^(n)P_3`.

Options

A. 5
B. 8
C. 6
D. 4

Answer

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

### Step 1: Write the Permutation Formula
The permutation formula is given by:
\[ nP_r = \frac{n!}{(n-r)!} \]
So, we can express \( nP_5 \) and \( nP_3 \) using this formula:
\[ nP_5 = \frac{n!}{(n-5)!} \]
\[ nP_3 = \frac{n!}{(n-3)!} \]

### Step 2: Substitute the Permutation Values into the Equation
Now, substitute these expressions into the original equation:
\[ \frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!} \]

### Step 3: Cancel \( n! \) from Both Sides
Since \( n! \) appears on both sides, we can cancel it out (assuming \( n! \neq 0 \)):
\[ \frac{1}{(n-5)!} = 20 \cdot \frac{1}{(n-3)!} \]

### Step 4: Rewrite the Factorials
Rearranging gives:
\[ (n-3)! = 20 \cdot (n-5)! \]
Now, we can express \( (n-3)! \) in terms of \( (n-5)! \):
\[ (n-3)(n-4)(n-5)! = 20 \cdot (n-5)! \]

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

### Step 6: Expand and Rearrange the Equation
Expanding the left side:
\[ n^2 - 7n + 12 = 20 \]
Now rearranging gives:
\[ n^2 - 7n - 8 = 0 \]

### Step 7: Factor the Quadratic Equation
Now we will factor the quadratic equation:
\[ (n - 8)(n + 1) = 0 \]

### Step 8: Solve for \( n \)
Setting each factor to zero gives:
1. \( n - 8 = 0 \) → \( n = 8 \)
2. \( n + 1 = 0 \) → \( n = -1 \)

Since \( n \) must be a positive integer, we take:
\[ n = 8 \]

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