Question

The number of integer solutions of the equation $x_1 + x_2 + x_3 + x_4 = 50$, where $x_1 \ge 1$, $x_2 \ge 2$, $x_3 \ge 0$, and $x_4 \ge 0$, is: 

• A. 19600
• B. 19200
• C. 20200
• D. 18400

Hint:

For problems involving the sum of variables with lower bounds, transform the variables to remove the constraints, and then use the stars and bars method to find the number of solutions.

Answer 1

The Correct Option is B

Given:

• Equation: \(x_1 + x_2 + x_3 + x_4 = 50\)
• Constraints: \(x_1 \ge 1\), \(x_2 \ge 2\), \(x_3 \ge 0\), \(x_4 \ge 0\)

Goal: Find the number of integer solutions. Steps:

1. Adjust variables: Let \(y_1 = x_1 - 1\), \(y_2 = x_2 - 2\), \(y_3 = x_3\), \(y_4 = x_4\). Then \(x_1 = y_1 + 1\), \(x_2 = y_2 + 2\), \(x_3 = y_3\), \(x_4 = y_4\). The constraints become \(y_1 \ge 0\), \(y_2 \ge 0\), \(y_3 \ge 0\), \(y_4 \ge 0\).
2. Substitute in the equation: \[ (y_1 + 1) + (y_2 + 2) + y_3 + y_4 = 50 \] \[ y_1 + y_2 + y_3 + y_4 = 50 - 1 - 2 \] \[ y_1 + y_2 + y_3 + y_4 = 47 \]
3. Use stars and bars method: We have 47 stars (representing 47 units) and 3 bars (to divide into 4 groups). The number of solutions is \(\binom{47 + 3}{3} = \binom{50}{3}\).
4. Calculate \(\binom{50}{3}\): \[ \binom{50}{3} = \frac{50!}{3! \cdot 47!} \] \[ = \frac{50 \cdot 49 \cdot 48}{3 \cdot 2 \cdot 1} \] \[ = \frac{50 \cdot 49 \cdot 48}{6} \] \[ = 50 \cdot 49 \cdot 8 \] \[ = 19600 \]

Answer: (a) 19600