Find n if `""^(n)P_(4)=18 ""^(n-1)P_3`

Options

A. 12
B. 8/5
C. 18
D. 54/17

Answer

To solve the equation \( nP_4 = 18 \cdot (n-1)P_3 \), we will follow these steps:

### Step 1: Write the permutation formulas
The permutation formula is given by:
\[ nP_r = \frac{n!}{(n-r)!} \]
Using this, we can express \( nP_4 \) and \( (n-1)P_3 \):
\[ nP_4 = \frac{n!}{(n-4)!} \]
\[ (n-1)P_3 = \frac{(n-1)!}{(n-1-3)!} = \frac{(n-1)!}{(n-4)!} \]

### Step 2: Substitute the permutation formulas into the equation
Now, substituting these into the original equation:
\[ \frac{n!}{(n-4)!} = 18 \cdot \frac{(n-1)!}{(n-4)!} \]

### Step 3: Cancel the common terms
Since both sides have \( (n-4)! \), we can cancel it out:
\[ n! = 18 \cdot (n-1)! \]

### Step 4: Rewrite \( n! \) in terms of \( (n-1)! \)
We know that:
\[ n! = n \cdot (n-1)! \]
So we can rewrite the equation as:
\[ n \cdot (n-1)! = 18 \cdot (n-1)! \]

### Step 5: Cancel \( (n-1)! \) from both sides
Assuming \( (n-1)! \neq 0 \) (which is true for \( n \geq 1 \)), we can cancel \( (n-1)! \):
\[ n = 18 \]

### Conclusion
Thus, the value of \( n \) is:
\[ \boxed{18} \]