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.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85f5be8c-91b2-4e3c-bb08-c2b9c395db43%2Fa900fd9b-abd6-4a45-8b60-020ebf0a4fcc%2Fti4ll9k_processed.png&w=3840&q=75)
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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