Let `f (n) = |{:(n,,n+1,,n+2),(.^(n)P_(n),,.^(n+1)P_(n+1),,.^(n+2)P_(n+2)),(.^(n)C_(n),,.^(n+1)C_(n+1),,.^(n+2)C_(n+2)):}|` where the sysmbols have their usual neanings .then f(n) is divisible by

Options

A. `n^(2)+n+1`
B. `(n+1)!`
C. `n!`
D. none of these

Explanation

To solve the problem, we need to evaluate the determinant given in the function \( f(n) \) and determine its divisibility. Let's break it down step by step.

### Step 1: Write the Determinant
We start with the determinant defined by the function:
\[ f(n) = \begin{vmatrix} n & n+1 & n+2 \\ nP_n & (n+1)P_{n+1} & (n+2)P_{n+2} \\ nC_n & (n+1)C_{n+1} & (n+2)C_{n+2} \end{vmatrix} \]

### Step 2: Substitute Values
Using the definitions of permutations and combinations:
- \( nP_n = n! \)
- \( (n+1)P_{n+1} = (n+1)! \)
- \( (n+2)P_{n+2} = (n+2)! \)
- \( nC_n = 1 \)
- \( (n+1)C_{n+1} = 1 \)
- \( (n+2)C_{n+2} = 1 \)

We can rewrite the determinant as:
\[ f(n) = \begin{vmatrix} n & n+1 & n+2 \\ n! & (n+1)! & (n+2)! \\ 1 & 1 & 1 \end{vmatrix} \]

### Step 3: Perform Row Operations
To simplify the determinant, we can perform column operations:
- Replace \( C_2 \) with \( C_2 - C_1 \)
- Replace \( C_3 \) with \( C_3 - C_1 \)

This gives us:
\[ f(n) = \begin{vmatrix} n & n+1 & n+2 \\ n! & (n+1)! - n! & (n+2)! - n! \\ 1 & 0 & 0 \end{vmatrix} \]

### Step 4: Simplify the Second Row
Calculating the second row:
- \( (n+1)! - n! = n! \cdot n \)
- \( (n+2)! - n! = (n+2)(n+1)n! - n! = (n^2 + 3n + 2 - 1)n! = (n^2 + 3n + 1)n! \)

Thus, we have:
\[ f(n) = \begin{vmatrix} n & n+1 & n+2 \\ n! & n \cdot n! & (n^2 + 3n + 1)n! \\ 1 & 0 & 0 \end{vmatrix} \]

### Step 5: Factor Out \( n! \)
Now we can factor \( n! \) out of the second row:
\[ f(n) = n! \begin{vmatrix} n & n+1 & n+2 \\ 1 & n & n^2 + 3n + 1 \\ 1 & 0 & 0 \end{vmatrix} \]

### Step 6: Expand the Determinant
Now we can expand this determinant:
\[ = n! \left( n \begin{vmatrix} n & n+2 \\ 1 & 0 \end{vmatrix} - (n+1) \begin{vmatrix} 1 & n+2 \\ 1 & 0 \end{vmatrix} \right) \]

Calculating the 2x2 determinants:
- \( \begin{vmatrix} n & n+2 \\ 1 & 0 \end{vmatrix} = -n - 2 \)
- \( \begin{vmatrix} 1 & n+2 \\ 1 & 0 \end{vmatrix} = - (n+2 - 1) = - (n+1) \)

Thus,
\[ f(n) = n! \left( n(-n - 2) + (n+1)(n+1) \right) \]

### Step 7: Final Simplification
This simplifies to:
\[ f(n) = n! \left( -n^2 - 2n + n^2 + 2n + 1 \right) = n! (n^2 + n + 1) \]

### Conclusion
Thus, we conclude that:
\[ f(n) = n! (n^2 + n + 1) \]
Therefore, \( f(n) \) is divisible by \( n! \) and \( n^2 + n + 1 \).