# Help Me To Find The Number Of Equation Solutions

Help me to solve this problem:

Find the number of non-negative integer solution of the equation:
$$5x_{1}+x_{2}+x_{3}+x_{4}+x_{5}=14$$ Note by Mashrur Fazla
5 years, 5 months ago

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If one number is thrice the other and their sum is 16 find the numbers

- 5 years, 5 months ago

It's the coefficient of z^14 in the generating function

$$G(z)=\dfrac{1}{(1-z^5)\,(1-z)^4}$$

$$[z^{14}]G(z)=\binom{17}{3}+\binom{12}{3}+\binom{7}{3} = 935$$

- 5 years, 5 months ago

This looks good. Can you please share a link which explains generating functions and their applications? I have often seen people using this.

Thanks!

- 5 years, 5 months ago

Generatingfunctionology is a great book for that, you can (legally) download the second edition of that book here.

- 5 years, 5 months ago

Tim, what a great guy, I'm glad that you respect copyrights. Thank you!

- 5 years, 5 months ago

the book is good thanks Tim

- 5 years, 5 months ago

You can learn some concepts form this article http://www.campusgate.co.in/2013/09/integer-solutions-using-coefficient.html

- 5 years, 5 months ago

Thanks for this link

- 5 years, 5 months ago

This file covers some of the interesting things that can be done using GF, and go through the references, which are good enough, I think.

- 5 years, 5 months ago

Thank u

- 5 years, 5 months ago

Case I x1= 0
x2 + x3 + x4 + x5 =14

Above is a linear Diophantine equation.

The number of non-negative solution to above equation is given by

(14+4-1)C(4-1) = 17C3 = 680

Case II x1= 1
x2 + x3 + x4 + x5 =9

The number of non-negative solution to above equation is = 12C3 = 220

Case III x1= 2
x2 + x3 + x4 + x5 =4

The number of non-negative solution to above equation is =7C3 = 35

Hence total number of non-negative integer solutions = 680 + 220 +30 = 935

- 5 years, 5 months ago