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Poopsilon

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**Comparing Partions of a Natural Number**

## Homework Statement

Let [itex]r(n)[/itex] denote the number of ordered triples of natural numbers [itex](a,b,c)[/itex] such that [itex]a + 2b + 3c = n[/itex], for [itex]n\geq 0[/itex]. Prove that this is equal to the number of ways of writing [itex]n = x + y + z[/itex] with [itex]0\leq x \leq y \leq z[/itex] for [itex]x,y,z[/itex] natural numbers.

## The Attempt at a Solution

I don't really have a whole lot of combinatorial tools at my disposal here. I proved earlier that r(n) = the integer nearest to [itex]\frac{(n+3)^2}{12}[/itex], but I don't think that's really going to help. I feel like I need to construct some 1-1 function between these two sets. But I can't find any way of uniquely associating a choice of (a,b,c) with an x≤y≤z or visa-versa.

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