The number of solutions of `x_1 + x_2 + x_3 = 51` (X1, X2, X3 being odd natural numbers) is

Options

A. 300
B. 325
C. 330
D. 350

Answer

To solve the problem of finding the number of solutions for the equation \( x_1 + x_2 + x_3 = 51 \) where \( x_1, x_2, x_3 \) are odd natural numbers, we can follow these steps:

### Step 1: Express odd natural numbers in terms of integers
Since \( x_1, x_2, x_3 \) are odd natural numbers, we can express them in the following form:
\[ x_1 = 2a + 1, \quad x_2 = 2b + 1, \quad x_3 = 2c + 1 \]
where \( a, b, c \) are non-negative integers.

### Step 2: Substitute into the equation
Substituting these expressions into the equation gives:
\[ (2a + 1) + (2b + 1) + (2c + 1) = 51 \]
This simplifies to:
\[ 2a + 2b + 2c + 3 = 51 \]

### Step 3: Simplify the equation
Now, we can simplify this equation:
\[ 2a + 2b + 2c = 51 - 3 \]
\[ 2a + 2b + 2c = 48 \]
Dividing the entire equation by 2, we get:
\[ a + b + c = 24 \]

### Step 4: Find the number of non-negative integer solutions
We need to find the number of non-negative integer solutions to the equation \( a + b + c = 24 \). This is a classic problem in combinatorics and can be solved using the "stars and bars" theorem.

According to the stars and bars theorem, the number of solutions for the equation \( x_1 + x_2 + \ldots + x_r = n \) in non-negative integers is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 24 \) and \( r = 3 \) (since we have three variables: \( a, b, c \)).

### Step 5: Apply the formula
Substituting the values into the formula gives:
\[ \binom{24 + 3 - 1}{3 - 1} = \binom{26}{2} \]

### Step 6: Calculate \( \binom{26}{2} \)
Now we calculate \( \binom{26}{2} \):
\[ \binom{26}{2} = \frac{26 \times 25}{2 \times 1} = \frac{650}{2} = 325 \]

### Conclusion
Thus, the number of solutions of \( x_1 + x_2 + x_3 = 51 \) where \( x_1, x_2, x_3 \) are odd natural numbers is:
\[ \boxed{325} \]