Question

The number of non negative integral solutions of the equation \( x+y+z+t=10 \) when \( x \ge 2, z \ge 5 \) is

• A. 80
• B. 20
• C. 50
• D. 10

Hint:

To handle constraints of the form \( x_i \ge k \), simply subtract the sum of all lower bounds \( k \) from the total sum \( N \) on the RHS. The new sum becomes \( N' = N - \sum k \). Then apply the standard formula \( \binom{N'+r-1}{r-1} \).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:

This is a combinatorics problem solvable using the "stars and bars" method (multinomial theorem). We first adjust the equation to remove the lower bound constraints.
Step 2: Key Formula or Approach:

The number of non-negative integral solutions to \( x_1 + x_2 + \dots + x_r = n \) is given by \( \binom{n+r-1}{r-1} \).
Step 3: Detailed Explanation:

Given equation: \[ x + y + z + t = 10 \] Constraints: \[ x \ge 2, \quad z \ge 5, \quad y \ge 0, \quad t \ge 0 \] Let's transform variables to handle constraints: Let \( x = x' + 2 \) where \( x' \ge 0 \). Let \( z = z' + 5 \) where \( z' \ge 0 \). Substitute these into the original equation: \[ (x' + 2) + y + (z' + 5) + t = 10 \] \[ x' + y + z' + t + 7 = 10 \] \[ x' + y + z' + t = 10 - 7 \] \[ x' + y + z' + t = 3 \] Now, we need to find the number of non-negative integral solutions for this new equation. Here, \( n = 3 \) (sum) and \( r = 4 \) (number of variables). Using the formula: \[ \text{Number of solutions} = \binom{n+r-1}{r-1} = \binom{3+4-1}{4-1} \] \[ = \binom{6}{3} \] \[ = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} \] \[ = 20 \]
Step 4: Final Answer:

The number of solutions is 20.