All the natural possibilities of x1 + x2 + x3 + x4 = 12 if at least one of these variables is equal or greater than 5?

[Kayton]

So, there is this system where the order matters, xk can be 0 and i would like to know all the possibilites for this system in which there is at least one variable equal or greater than 5

 

[Dr.Peterson]

So, there is this system where the order matters, xk can be 0 and i would like to know all the possibilites for this system in which there is at least one variable equal or greater than 5

Do you mean you want to list them, or count them?

This is a problem in restricted ordered 4-partitions (or something like that -- I've seen different terminology), which will not have a simple formula due to the restriction. You can count all ordered partitions using "stars and bars", but the restriction makes it harder.

 

[pka]

So, there is this system where the order matters, xk can be 0 and i would like to know all the possibilites for this system in which there is at least one variable equal or greater than 5

I assume that there are four vairable \(x_1+x_2+x_3+x_4=12\) where at least one \(x_k\ge 5\).
I agree with Prof. Peterson, that last condition complicates matters. In fact I had to go to generating polynomials to solve.
It is well known that there are \(\dbinom{N+J-1}{N}=\dfrac{(N+J-1)!}{N!\cdot(J-1)!}\) ways to put \(N\) identical objects into \(J\) distinct cells.
Applied to this question \(\dbinom{12+4-1}{12}=455\): SEE HERE
To see the application of generating polynomials SEE HERE seeing the term \(455x^{12}\) agreeing with the other approach.
Of course those four hundred and fifty-five solutions many do not have a solution in which one variable is at least five.
Again turning to generating polynomials SEE HERE, we see the term \(35x^{12}\) telling us that there are thirty-five ways that one of those solutions is at least five. Here is the list of \(35\) solutions having no solution more than four.
\(\begin{array}{*{20}{c}} {1,3,4,4}&{\enclose{box}{12}} \\ {2,3,3,4}&{\enclose{box}{12}} \\ {0,4,4,4}&{\enclose{box}{4}} \\ {3,3,3,3}&{\enclose{box}{1}} \\
{2,2,4,4}&{\enclose{box}{6}} \end{array}\)
The first column contains the possible solutions. the second column tell us the number of ways those can be rearranged.
The sum of the second column is thirty-five.
So \(455-35=420\) is your answer.