MHB Not an exercise, more of a question.

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To determine the unique ways to sum to a number n using u addends, where order matters and all addends are positive integers, a combinatorial approach is necessary. The problem can be framed as finding the number of integer solutions to the equation x1 + x2 + ... + xu = n, with the constraint that each xi > 0. The generating functions or the stars and bars theorem can be applied to simplify the calculations. For the example of n=6 with u=4, the formula involves calculating permutations of the partitions of n into u parts. This method avoids the tediousness of manually listing permutations and ensures all unique combinations are counted.
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If I have a number n and I want to know all the unique ways in which I can use u addends to get that number... How do I do it?

For example: If the number is 6 and I want to see how many unique ways I can add up to it (order matters: $$1 + 1 + 2 + 2\ne 2 + 2 + 1 + 1$$) by using 4 addends, what's the formula? (This is, in fact, considering that all the addends are greater than 0 and whole numbers.)

I can do it by looking for each unique way to add up to n using u addends and then factoring in the number of ways each can be permutated, including repeated numbers, but that seems too tedious.
 
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I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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