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2

The two things you wanted to know: THE MISTAKE: $x \not= y \land y \not= z \not \implies x \not =z$ Take any counter-example, as in Parth Kohli's comment. $(x,y)$ is an ordered pair, so essentially $(x,y) \in \mathbb{N} \times \mathbb{N}$.


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A simple counter example will suffice. Take $x=z=1$, $y=2$. Then $x \neq y$ and $y\neq z$, but $x = z$. As to your last question, we use $x,y \in \mathbb{N}$ to mean that both $x$ and $y$ are natural numbers. This differs from $(x, y) \in \mathbb{N^2}$ where $(x,y)$ is an ordered pair.


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There are many proofs, most of them are elementary. For example: non-intersecting lattice paths and LGV-lemma — see e.g. Bressoud. Proofs and Confirmations (ch. 3) RSK-correspondence — see e.g. Stanley. Enumerative combinatorics (vol. 2, 7.20) counting lozenge tilings using condensation


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Any $n \geq 2$. Partition according to parity so that $A = A_n$ and $B = S_n - A_n$. Take $\pi = (1, 2)$, which is an odd permutation. Since any even permutation in $A$ multiplied by the odd permutation $\pi$ will result in an odd permutation in $B$, we are done.


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This is the number of partitions of a given integer into a sum. Say, $$m = 6,$$ and we would like to write it as the sum of 3 positive summands. To do this, write $$ 6 = 1 + 1 + 1 + 1 + 1 + 1. $$ A sum is formed by choosing 2 of the 5 pluses in the above expression; e.g, $$ 1 + 1 +' 1 + 1 + 1 +' 1 = 2 + 3 + 1. $$ The number of ways to choose 2 pluses is "5 ...


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Partition of a set $X$ is a collection of pairwise disjoint subsets of $X$ such that union of all these subsets gives $X$. Any partition naturally sets an equivalence relation and vise versa. The partition in your question can be set by equivalence relation that can be described in many ways. For example like this: we say that $x$ is equivalent to $y$ if $x$ ...


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Every couple of fractions $c/a$ and $d/b$ in the Stern-Brocot tree can be represented (in the inverse notation according to Concrete Mathematics) by the matrix ${\bf M} = \left\| {\,\begin{array}{*{20}c} a & b \\ c & d \\ \end{array}\,} \right\|$, where, iff the fractions are the generators of another fraction in the tree, then the ...



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