$x_1 + x_2 + x_3 + x_4 = 32$ 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 ${28 \choose 3}-1$.
Best Answer
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.