The no. of integral solutions of `x_1+x_2+x_3=0` with `x_k, >=-5` is

Options

A. `""^(15)C_(2)`
B. `""^(16)C_(2)`
C. `""^(17)C_(2)`
D. `""^(18)C_(2)`

Answer

To find the number of integral solutions for the equation \( x_1 + x_2 + x_3 = 0 \) with the constraint \( x_k \geq -5 \) for \( k = 1, 2, 3 \), we can follow these steps:

### Step 1: Transform the Variables
Since \( x_k \geq -5 \), we can make a substitution to simplify the problem. Let:
\[ y_1 = x_1 + 5, \quad y_2 = x_2 + 5, \quad y_3 = x_3 + 5 \]
This transformation ensures that \( y_k \geq 0 \) for \( k = 1, 2, 3 \).

### Step 2: Rewrite the Original Equation
Substituting \( x_k \) in terms of \( y_k \) into the original equation gives:
\[ (y_1 - 5) + (y_2 - 5) + (y_3 - 5) = 0 \]
Simplifying this, we have:
\[ y_1 + y_2 + y_3 - 15 = 0 \]
Thus, we can rewrite it as:
\[ y_1 + y_2 + y_3 = 15 \]

### Step 3: Apply the Stars and Bars Theorem
Now, we need to find the number of non-negative integral solutions to the equation \( y_1 + y_2 + y_3 = 15 \). This is a classic application of the "stars and bars" theorem, which states that the number of ways to distribute \( n \) indistinguishable objects (stars) into \( r \) distinguishable boxes (variables) is given by:
\[ \binom{n + r - 1}{r - 1} \]
In our case, \( n = 15 \) (the total sum we want) and \( r = 3 \) (the number of variables). Therefore, we have:
\[ \text{Number of solutions} = \binom{15 + 3 - 1}{3 - 1} = \binom{17}{2} \]

### Step 4: Calculate the Binomial Coefficient
Now, we calculate \( \binom{17}{2} \):
\[ \binom{17}{2} = \frac{17 \times 16}{2 \times 1} = \frac{272}{2} = 136 \]

### Conclusion
Thus, the number of integral solutions to the equation \( x_1 + x_2 + x_3 = 0 \) with the constraint \( x_k \geq -5 \) is:
\[ \boxed{136} \]