Stars and Bars with bounds but order matters

Question

This question is related to Stars and Bars with bounds

I have what I thought is a fairly simple problem: Count non-negative integer solutions to the equation

$$x_1 + x_2 + x_3 + x_4 + x_5 = 23$$

such that $0 \leq x_1 \leq 9$.

The accepted answer uses nCr such that order does not matter. But if order does matter can the function be replaced from \begin{align*} [x^j](1+x)^n=\binom{n}{j} \end{align*} to then be

\begin{align*} [x^j](1+x)^n=n^j \end{align*}

$x^j^n=\binom{n}{j}$ is just an example of the $[x^j]$ notation, it's not actually related to the answer. — Misha Lavrov, Feb 13 '21 at 21:01
Okay I see, so how does \binom{k+4}{4}x^k translate if order matters? — Coffee_Overflow, Feb 13 '21 at 21:12
What exactly do you mean in this context by order matters? — Brian M. Scott, Feb 13 '21 at 21:13
@BrianM.Scott The "bin" order. The original treats 1+2+3+4+5 as a series such that their order doesnt matter. AKA 9+9+5+0+0 is that same as 0+0+5+9+9. What if I want to count these two possibilities as unique solutions? — Coffee_Overflow, Feb 13 '21 at 21:17
You are mistaken. $9+9+5+0+0$ and $0+0+5+9+9$ are counted in the original as two different solutions. — user, Feb 13 '21 at 21:21
@Coffee_Overflow: If one actually uses a combination of stars and bars and, to deal with the upper bound, inclusion-exclusion, order of the bins is automatically taken into account. — Brian M. Scott, Feb 13 '21 at 21:21

score 0 · Answer 1 · answered Feb 13 '21 at 22:04

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My confusion was with why the "Choose" function was used rather than the "Permute" function when order matters. According to @user by the nature of the stars and bars and by extension, generating function it takes the order of the bins into account despite using the choose function as a basis.

This offers the best example for how this happens: https://youtu.be/zjtxJSkjejc?t=308

answered Feb 13 '21 at 22:04

Coffee_Overflow

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As already pointed out in another comment the "Choose" function was used only as example to demonstrate what is the action of the coefficient extractor. It is absolutely hopeless trying to modify a method without its understanding. – user Feb 14 '21 at 00:19

Stars and Bars with bounds but order matters

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