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$x_1 + x_2 + x_3 + x_4 = 32$ and $x_1 > x_2 > x_3 > 0$ and $0<x_4\le25$. There are how many solutions for this equation?

I was trying to enumerate the solutions one-by-one but that didn't work

my try: so from this $x_1 > x_2 > x_3 > 0$ we have $x_3\ge1 , x_2\ge2, x_1\ge3$ and from $0<x_4\le25$ we have $x_4\ge1$ so we can write : $(x_1-3)+(x_2-2)+(x_3-1)+(x_4-1)=32-3-2-1-1=25$. then ${25+3 \choose 3}={28 \choose 3}$. because of $x_4\le25$ we have ${28 \choose 3}-1$.

N. F. Taussig
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Elsa Hejazian
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    writing all of the solutions one by one ... cant find any proper way to solve it – Elsa Hejazian May 08 '21 at 08:32
  • what is the order with $x_4? just on $x_1>x_2>x_3?$ – tryst with freedom May 08 '21 at 08:40
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    $x_4$ doesn't have order – Elsa Hejazian May 08 '21 at 08:41
  • We use \le for $\le$, not the digraph $<=$. Also you should have more substantiative work on your paper; please edit that into the question. Every little detail matters. – Parcly Taxel May 08 '21 at 08:44
  • Look for "generating functions", there are many examples on MSE, like this one. – rtybase May 08 '21 at 08:53
  • There is a simplification you can make. The inequalities suggest $x_3$ must always be at least 1, $x_2$ be at least 2 and $x_1$ at least 3. Thus you can consider instead $$z_1+z_2+z_3+z_4=25$$ instead where the $z$'s (other than $z_4\leq 24$) now have no restriction other than nonnegative. But now you can rule out the one solution where $z_4>24$ by hand. – Ninad Munshi May 08 '21 at 08:53
  • Consider fixing $x_4$ and try solving the equation $x_1+x_2+x_3=32-x_4$ (taking $x_4$ as a parameter), then sum the number of solutions up. – atzlt May 08 '21 at 09:28
  • I don't understand the point in continuing these discussions, clearly nobody looks at the links posted, here is another one and I vote to close this questions, since OP didn't update it with anything they tried (not just text "I tried"). – rtybase May 08 '21 at 09:31
  • @ElsaHejazian I have posted 2 links, both containing a technique to replace $$x_1=x_2+x_1^{'}, x_1^{'}>0$$ $$x_2=x_3+x_2^{'}, x_2^{'}>0$$ Substitute and you will have a new equation with $x_1^{'}>0,x_2^{'}>0,x_3>0$ and $0<x_4<25$. – rtybase May 08 '21 at 10:12
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    This is not linear algebra, please use adequate tags – jjagmath May 08 '21 at 10:12
  • I never said this was stars and bars. All you have to do is treat the first bins as indistinguishable from each other, that way the bin with the smallest number is always the "third bin" and so on. Then subtract off the times the bins are equivalent. – Ninad Munshi May 08 '21 at 13:08

1 Answers1

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Quesiton says that $x_1,x_2,x_3$ are positive integers and greater than one another.Lets say that $x_2=x_1+ a$ where $a$ is positive integer ,as well in order to satisfy the order. Then , $x_3=x_2 +b$ where $b$ is positive integer by the same reason. So , $x_3=x_1+a+b$.

As a result , $x_1+x_2+x_3=3x_1+2a+b$ where $x_1,a,b$ are positive integers. Then , their generating functions can be written as $\frac{x^3}{1-x^3} \times \frac{x^2}{1-x^2} \times \frac{x}{1-x} $.

Lets come to $x_4$ , it is in the interval of $(0,25]$ , then its generating functions can be written $\frac{1-x^{26}}{1-x} - 1$.

The rest is given to you for calculating to find $[x^{32}]$. You can do it by hand but it is very cumbersome process, so i recommend you to use wolframalpha.

$\color{green}{Calculation}$:https://www.wolframalpha.com/input/?i=expanded+form+of+%28%28x%5E6%29*%28x-x%5E26%29%29%2F%28%281-x%5E3%29%281-x%5E2%29%281-x%29%5E2%29

Answer = $\color{red}{639}$

$\color{red}{NOTE}$= Question says $x_1>x_2>x_3$ but i solved it as to $x_3 > x_2 > x_1$ . The result is the same , nothing changes.

Not a Salmon Fish
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