If `n! = ((n + 4)!)/((n + 1)!)` , then the value of n is :

Options

A. 5
B. 18
C. 6
D. 9

Answer

To solve the equation \( n! = \frac{(n + 4)!}{(n + 1)!} \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ n! = \frac{(n + 4)!}{(n + 1)!} \]

### Step 2: Expand the factorials
We can expand \( (n + 4)! \) as follows:
\[ (n + 4)! = (n + 4)(n + 3)(n + 2)(n + 1)(n!) \]
Thus, we can rewrite the right side of the equation:
\[ \frac{(n + 4)!}{(n + 1)!} = \frac{(n + 4)(n + 3)(n + 2)(n + 1)(n!)}{(n + 1)!} \]
Now, \( (n + 1)! = (n + 1)(n)(n - 1)(n - 2)...(1) \), so we can simplify:
\[ \frac{(n + 4)(n + 3)(n + 2)(n!)}{(n + 1)(n!)} = (n + 4)(n + 3)(n + 2) \]

### Step 3: Set the equation
Now we have:
\[ n! = (n + 4)(n + 3)(n + 2) \]

### Step 4: Test integer values for n
We will test integer values for \( n \) to find a solution.

- **For \( n = 1 \)**:
\[ 1! = 1 \quad \text{and} \quad (1 + 4)(1 + 3)(1 + 2) = 5 \cdot 4 \cdot 3 = 60 \quad \text{(not equal)} \]

- **For \( n = 2 \)**:
\[ 2! = 2 \quad \text{and} \quad (2 + 4)(2 + 3)(2 + 2) = 6 \cdot 5 \cdot 4 = 120 \quad \text{(not equal)} \]

- **For \( n = 3 \)**:
\[ 3! = 6 \quad \text{and} \quad (3 + 4)(3 + 3)(3 + 2) = 7 \cdot 6 \cdot 5 = 210 \quad \text{(not equal)} \]

- **For \( n = 4 \)**:
\[ 4! = 24 \quad \text{and} \quad (4 + 4)(4 + 3)(4 + 2) = 8 \cdot 7 \cdot 6 = 336 \quad \text{(not equal)} \]

- **For \( n = 5 \)**:
\[ 5! = 120 \quad \text{and} \quad (5 + 4)(5 + 3)(5 + 2) = 9 \cdot 8 \cdot 7 = 504 \quad \text{(not equal)} \]

- **For \( n = 6 \)**:
\[ 6! = 720 \quad \text{and} \quad (6 + 4)(6 + 3)(6 + 2) = 10 \cdot 9 \cdot 8 = 720 \quad \text{(equal)} \]

### Conclusion
The value of \( n \) that satisfies the equation is:
\[ \boxed{6} \]