If the number of positive integral solutions of `u+v+w=n` be denoted by `P_(n)` then the absolute value of `|{:(P_(n),P_(n+1),P_(n+2)),(P_(n+1),P_(n+2),P_(n+3)),(P_(n+2),P_(n+3),P_(n+4)):}|` is

Options

A. `-1`
B. `2`
C. `3`
D. `4`

Explanation

To solve the problem, we need to find the absolute value of the determinant given by:

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

### Step 1: Determine \( P(n) \)

The number of positive integral solutions of the equation \( u + v + w = n \) can be represented as \( P(n) \). Since \( u, v, w \) must all be at least 1, we can make a substitution:

Let \( u' = u - 1 \), \( v' = v - 1 \), and \( w' = w - 1 \). Then, the equation becomes:

\[ u' + v' + w' = n - 3 \]

where \( u', v', w' \geq 0 \). The number of non-negative integral solutions to this equation is given by the "stars and bars" theorem:

\[ P(n) = \binom{n-1}{2} \]

### Step 2: Calculate \( P(n+1), P(n+2), P(n+3), P(n+4) \)

Using the formula derived above, we can calculate:

- \( P(n+1) = \binom{n}{2} \)
- \( P(n+2) = \binom{n+1}{2} \)
- \( P(n+3) = \binom{n+2}{2} \)
- \( P(n+4) = \binom{n+3}{2} \)

### Step 3: Substitute into the Determinant

Now we substitute these values into the determinant \( D \):

\[ D = \begin{vmatrix} \binom{n-1}{2} & \binom{n}{2} & \binom{n+1}{2} \\ \binom{n}{2} & \binom{n+1}{2} & \binom{n+2}{2} \\ \binom{n+1}{2} & \binom{n+2}{2} & \binom{n+3}{2} \end{vmatrix} \]

### Step 4: Calculate the Determinant

To calculate this determinant, we can use properties of determinants. Notice that the rows of the determinant are formed by consecutive binomial coefficients. This structure allows us to apply the determinant property that states if two rows (or columns) are identical, the determinant is zero.

In this case, we can see that:

\[ D = 0 \]

### Step 5: Find the Absolute Value

Since the determinant \( D \) is zero, the absolute value is:

\[ |D| = |0| = 0 \]

### Final Answer

Thus, the absolute value of the determinant is:

\[ \boxed{0} \]