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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.

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  • $\begingroup$ Try this as an example. $\endgroup$ – rtybase Jun 28 at 19:34
  • $\begingroup$ What about $x_4$ and $x_5$? Are they arbitrary? $\endgroup$ – user Jun 28 at 21:12
  • $\begingroup$ @user yes ,they are $\endgroup$ – LEO Jun 28 at 21:25
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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.

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  • $\begingroup$ thank you very much sir.By the way, is there any way to solve it using inclusion -exclusion without using generating functions $\endgroup$ – LEO Jun 28 at 22:37
  • $\begingroup$ At the moment I see no simple way to use inclusion-exclusion. $\endgroup$ – user Jun 29 at 6:12

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