Let $N$ be some positive integer, and consider finite sequences of natural numbers $x,y$ whose sum equals $N$.
We can in some cases build a total order among these sequences, provided we identify sequences differing only by the ordering of the elements.
For example, for $N=2$ we have $(1,1)\prec(2,0)$. Similarly, for $N=3$, we have $$(1,1,1)\prec(2,1,0) \prec (3,0,0).$$ For $N=4$, we have $$(1111)\prec(2110)\prec (2200)\prec (3100)\prec(4000).$$ Again, for $N=5$ we have the sequence: $$(11111)\prec(21110)\prec (22100)\prec (31100)\prec(32000)\prec(41000)\prec(50000).$$ How can we show that this always works (if it does)? More precisely, how do we show that the set of all non-increasing tuples of $N$ elements between $0$ and $N$ can be totally ordered via majorization?
This fails for $N=6$. The sequences $(3,1,1,1)$ and $(2,2,2,0)$ are incomparable, as $$3>2,$$ $$3+1+1<2+2+2.$$ In fact, all $N\geq6$ fail, since both $$(3,\underbrace{1,1,\ldots,1}_{N-3\text{ 1s}}),$$ $$(2,2,2,\underbrace{1,1,\ldots,1}_{N-6\text{ 1s}},0)$$ are incomparable too.