If `(n+1)! =12 xx [(n-1)!]`, find the value of n.

Explanation

To solve the equation \((n+1)! = 12 \times (n-1)!\), we can follow these steps:

### Step 1: Rewrite the factorial expression
We start with the equation:
\[ (n+1)! = 12 \times (n-1)! \]
We know that \((n+1)! = (n+1) \times n \times (n-1)!\). So we can rewrite the equation as:
\[ (n+1) \times n \times (n-1)! = 12 \times (n-1)! \]

### Step 2: Cancel out \((n-1)!\)
Since \((n-1)!\) appears on both sides of the equation, we can safely cancel it out (assuming \(n \neq 1\) since \(0! = 1\)):
\[ (n+1) \times n = 12 \]

### Step 3: Expand and rearrange the equation
Now we expand the left-hand side:
\[ n^2 + n = 12 \]
Rearranging gives us:
\[ n^2 + n - 12 = 0 \]

### Step 4: Factor the quadratic equation
Next, we need to factor the quadratic equation. We look for two numbers that multiply to \(-12\) and add to \(1\). The numbers \(4\) and \(-3\) work:
\[ (n + 4)(n - 3) = 0 \]

### Step 5: Solve for \(n\)
Setting each factor to zero gives us:
\[ n + 4 = 0 \quad \Rightarrow \quad n = -4 \quad \text{(not valid since } n \text{ must be positive)} \]
\[ n - 3 = 0 \quad \Rightarrow \quad n = 3 \]

### Conclusion
The only valid solution is:
\[ \boxed{3} \]