How many non-negative integer are there for $x_1+x_2+x_3+x_4+x_5=12$ where $x_1

Question

I have a combinatorics problem:

How many non-negative integer are there for $x_1+x_2+x_3+x_4+x_5=12$ where $x_1<x_2<x_3$

I tried to subtract the number of integers in which $x_1=x_2=x_3$ from the number of the whole possible solutions.However I could not do the rest.Can you enlighten me ,please?

Note: $x_1,x_2,x_3$ do not have to be consecutive. They are just smaller than each other ,respectively.

What about $x_4$ and $x_5$? Are they arbitrary? – user Jun 28 '20 at 21:12 — user, Jun 28 '20 at 21:12
@user yes ,they are – Not a Salmon Fish Jun 28 '20 at 21:25 — Not a Salmon Fish, Jun 28 '20 at 21:25

score 2 · Accepted Answer · answered Jun 28 '20 at 22:08

2

Let us apply the approach suggested in comment.

Let $x_2=x_1+a, x_3=x_2+b$ so that the original equality can be written as: $$ 3x_1+2a+b+x_4+x_5=12 $$ where $a$ and $b$ are positive and the other are non-negative integer numbers.

The number of integer solutions to the equation is: $$ [x^{12}]\frac1{1-x^3}\frac{x^2}{1-x^2}\frac{x}{1-x}\frac{1}{1-x}\frac{1}{1-x}= [x^{12}]\frac{x^3}{(1-x^3)(1-x^2)(1-x)^3}=189. $$ where $[x^n]$ means the coefficient at $x^n$ in the series expansion of the following expression.

answered Jun 28 '20 at 22:08

user

26,272

thank you very much sir.By the way, is there any way to solve it using inclusion -exclusion without using generating functions – Not a Salmon Fish Jun 28 '20 at 22:37
At the moment I see no simple way to use inclusion-exclusion. – user Jun 29 '20 at 06:12

How many non-negative integer are there for $x_1+x_2+x_3+x_4+x_5=12$ where $x_1

1 Answers1