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Consider the problem of finding the number of integral solutions of the equation ₁+X2+X3+ *4 = 20 for which x₁ ≥ 3, x2 ≥ 1, x3 ≥ 0 and £4 ≥ 5. - (a) Do the problem by introducing the variables y₁ = x₁ -3, y2 = x2 − 1, y3 = x3 and Y4 = x4 - 5. (b) Do the problem using generating functions.

Consider the problem of finding the number of integral solutions of the equation ₁+X2+X3+ *4 = 20 for which x₁ ≥ 3, x2 ≥ 1, x3 ≥ 0 and £4 ≥ 5. - (a) Do the problem by introducing the variables y₁ = x₁ -3, y2 = x2 − 1, y3 = x3 and Y4 = x4 - 5. (b) Do the problem using generating functions.

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter4: Polynomial And Rational Functions
Section4.4: Complex And Rational Zeros Of Polynomials
Problem 46E
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### Integral Solutions of Linear Diophantine Equations #### Problem Statement Consider the problem of finding the number of integral solutions of the equation: \[x_1 + x_2 + x_3 + x_4 = 20\] for which: \[x_1 \geq 3, \quad x_2 \geq 1, \quad x_3 \geq 0, \quad \text{and} \quad x_4 \geq 5\] #### Solution Approaches ##### (a) Variable Transformation Method Do the problem by introducing the variables: \[ y_1 = x_1 - 3, \quad y_2 = x_2 - 1, \quad y_3 = x_3, \quad \text{and} \quad y_4 = x_4 - 5 \] ##### (b) Generating Functions Method Do the problem using generating functions. --- For educational purposes, let us briefly explain both methods. --- #### (a) Variable Transformation Method By substituting the variables as given: - \( y_1 = x_1 - 3 \implies x_1 = y_1 + 3 \) - \( y_2 = x_2 - 1 \implies x_2 = y_2 + 1 \) - \( y_3 = x_3 \) - \( y_4 = x_4 - 5 \implies x_4 = y_4 + 5 \) Substitute these into the original equation: \[ (y_1 + 3) + (y_2 + 1) + y_3 + (y_4 + 5) = 20 \] Simplify the equation: \[ y_1 + y_2 + y_3 + y_4 + 9 = 20 \] \[ y_1 + y_2 + y_3 + y_4 = 11 \] Now, \(y_1, y_2, y_3, y_4 \geq 0\) are non-negative integers. The problem now reduces to finding the number of non-negative integer solutions of the above equation, which is a classic stars and bars problem. The number of solutions is given by the binomial coefficient: \[ \binom{11 + 4 - 1
Transcribed Image Text:### Integral Solutions of Linear Diophantine Equations #### Problem Statement Consider the problem of finding the number of integral solutions of the equation: \[x_1 + x_2 + x_3 + x_4 = 20\] for which: \[x_1 \geq 3, \quad x_2 \geq 1, \quad x_3 \geq 0, \quad \text{and} \quad x_4 \geq 5\] #### Solution Approaches ##### (a) Variable Transformation Method Do the problem by introducing the variables: \[ y_1 = x_1 - 3, \quad y_2 = x_2 - 1, \quad y_3 = x_3, \quad \text{and} \quad y_4 = x_4 - 5 \] ##### (b) Generating Functions Method Do the problem using generating functions. --- For educational purposes, let us briefly explain both methods. --- #### (a) Variable Transformation Method By substituting the variables as given: - \( y_1 = x_1 - 3 \implies x_1 = y_1 + 3 \) - \( y_2 = x_2 - 1 \implies x_2 = y_2 + 1 \) - \( y_3 = x_3 \) - \( y_4 = x_4 - 5 \implies x_4 = y_4 + 5 \) Substitute these into the original equation: \[ (y_1 + 3) + (y_2 + 1) + y_3 + (y_4 + 5) = 20 \] Simplify the equation: \[ y_1 + y_2 + y_3 + y_4 + 9 = 20 \] \[ y_1 + y_2 + y_3 + y_4 = 11 \] Now, \(y_1, y_2, y_3, y_4 \geq 0\) are non-negative integers. The problem now reduces to finding the number of non-negative integer solutions of the above equation, which is a classic stars and bars problem. The number of solutions is given by the binomial coefficient: \[ \binom{11 + 4 - 1
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