If x is a positive integer, then `|(x!,(x +1)!,(x +2)!),((x +1)!,(x +2)!,(x +3)!),((x +2)!,(x +3)!,(x +4)!)|` is equal to

Options

A. `2x! (x +1) !`
B. `2x! (x +1) ! (x +2)!`
C. `2x ! (x +3)!`
D. `2(x +1) ! (x +2) ! (x _3)!`

Explanation

To solve the determinant

\[ D = \begin{vmatrix} x! & (x+1)! & (x+2)! \\ (x+1)! & (x+2)! & (x+3)! \\ (x+2)! & (x+3)! & (x+4)! \end{vmatrix} \]

we can simplify the factorials in the determinant.

### Step 1: Factor out common terms from each row

1. **First Row**: Factor out \(x!\):
\[ x! \begin{vmatrix} 1 & x+1 & (x+2) \\ (x+1)! & (x+2)! & (x+3)! \\ (x+2)! & (x+3)! & (x+4)! \end{vmatrix} \]

2. **Second Row**: Factor out \((x+1)!\):
\[ (x+1)! \begin{vmatrix} 1 & 1 & (x+2) \\ 1 & (x+2) & (x+3) \\ (x+2)! & (x+3)! & (x+4)! \end{vmatrix} \]

3. **Third Row**: Factor out \((x+2)!\):
\[ (x+2)! \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & (x+3) \\ 1 & (x+3) & (x+4) \end{vmatrix} \]

Now, we have:
\[ D = x! (x+1)! (x+2)! \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & (x+3) \\ 1 & (x+3) & (x+4) \end{vmatrix} \]

### Step 2: Evaluate the 3x3 determinant

Now we will evaluate the determinant:
\[ \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & (x+3) \\ 1 & (x+3) & (x+4) \end{vmatrix} \]

Using the determinant formula, we can expand this determinant:
\[ = 1 \cdot \begin{vmatrix} 1 & (x+3) \\ (x+3) & (x+4) \end{vmatrix} - 1 \cdot \begin{vmatrix} 1 & (x+3) \\ 1 & (x+4) \end{vmatrix} + 1 \cdot \begin{vmatrix} 1 & 1 \\ 1 & (x+3) \end{vmatrix} \]

Calculating each of these 2x2 determinants:
1. \(\begin{vmatrix} 1 & (x+3) \\ (x+3) & (x+4) \end{vmatrix} = 1 \cdot (x+4) - (x+3) \cdot (x+3) = x + 4 - (x^2 + 6x + 9) = -x^2 - 5x - 5\)

2. \(\begin{vmatrix} 1 & (x+3) \\ 1 & (x+4) \end{vmatrix} = 1 \cdot (x+4) - (x+3) \cdot 1 = x + 4 - (x + 3) = 1\)

3. \(\begin{vmatrix} 1 & 1 \\ 1 & (x+3) \end{vmatrix} = 1 \cdot (x+3) - 1 \cdot 1 = x + 3 - 1 = x + 2\)

Putting it all together:
\[ = -x^2 - 5x - 5 - 1 + (x + 2) = -x^2 - 4x - 4 \]

### Step 3: Final Calculation

Now substituting back into our expression for \(D\):
\[ D = x! (x+1)! (x+2)! (-x^2 - 4x - 4) \]

### Conclusion

Thus, the final result is:
\[ D = -x! (x+1)! (x+2)! (x^2 + 4x + 4) \]