The number of positive integral solutions of `x+y+z=n,n in N, n >= 3` is

Options

A. `""^(n-1)C_(2)`
B. `""^(n-1)P_(2)`
C. `n(n-1)`
D. `n^(2)`

Answer

To find the number of positive integral solutions of the equation \( x + y + z = n \) where \( n \) is a natural number and \( n \geq 3 \), we can follow these steps:

### Step 1: Understanding the Problem
We need to find the number of positive integral solutions for the equation \( x + y + z = n \). Since \( x, y, z \) must all be positive integers, we can rewrite the equation in terms of new variables.

### Step 2: Transforming the Variables
Let’s define new variables:
- \( x' = x - 1 \)
- \( y' = y - 1 \)
- \( z' = z - 1 \)

Since \( x, y, z \) are positive integers, \( x', y', z' \) will be non-negative integers (i.e., \( x', y', z' \geq 0 \)). The equation now becomes:
\[ (x' + 1) + (y' + 1) + (z' + 1) = n \]
This simplifies to:
\[ x' + y' + z' = n - 3 \]

### Step 3: Counting Non-negative Solutions
Now, we need to count the number of non-negative integral solutions to the equation \( x' + y' + z' = n - 3 \).

### Step 4: Applying the Stars and Bars Theorem
The number of non-negative integral solutions to the equation \( x_1 + x_2 + \ldots + x_k = r \) is given by the formula:
\[ \binom{r + k - 1}{k - 1} \]
In our case, \( r = n - 3 \) and \( k = 3 \) (since we have three variables: \( x', y', z' \)). Thus, we have:
\[ \text{Number of solutions} = \binom{(n - 3) + 3 - 1}{3 - 1} = \binom{n - 1}{2} \]

### Step 5: Final Result
Therefore, the number of positive integral solutions of the equation \( x + y + z = n \) is:
\[ \binom{n - 1}{2} \]