The value of n, if `""^(n)P_(5) = 20.""^(n)P_(3)` , is

Options

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

Answer

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

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

### Step 2: Set up the equation
Using the formula, we can write:
\[ nP_5 = \frac{n!}{(n-5)!} \quad \text{and} \quad nP_3 = \frac{n!}{(n-3)!} \]
Substituting these into the original equation gives us:
\[ \frac{n!}{(n-5)!} = 20 \cdot \frac{n!}{(n-3)!} \]

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

### Step 4: Rewrite the factorials
We can express \( (n-3)! \) in terms of \( (n-5)! \):
\[ (n-3)(n-4)(n-5)! = 20 \cdot (n-5)! \]
Now, we can cancel \( (n-5)! \) from both sides:
\[ (n-3)(n-4) = 20 \]

### Step 5: Expand and rearrange the equation
Expanding the left side gives:
\[ n^2 - 7n + 12 = 20 \]
Rearranging this leads to:
\[ n^2 - 7n - 8 = 0 \]

### Step 6: Factor the quadratic equation
Now we will factor the quadratic equation:
\[ (n - 8)(n + 1) = 0 \]
Setting each factor to zero gives us:
\[ n - 8 = 0 \quad \text{or} \quad n + 1 = 0 \]
Thus, we find:
\[ n = 8 \quad \text{or} \quad n = -1 \]

### Step 7: Determine the valid solution
Since \( n \) must be a non-negative integer (as it represents the number of items), we discard \( n = -1 \) and keep:
\[ n = 8 \]

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