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# A permutation problem

I have got this intriguing problem from my friend.

2 distinct permutations $$(x_1,x_2,...,x_n)$$ and $$(y_1,y_2,...,y_n)$$ of the set $$A=(1,2,...,n)$$ are considered to have T property if and only if $$x_i\neq y_i$$ for all $$i=1,2,...,n$$.

Prove that in $$(n-1)!+1$$ distinct permutations of the set A,we always find out 2 permutations that have T property.

Note by Hunter Killer
4 years ago

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Let $$P_i$$ be the permutation of $$A$$ such that its first element is $$1$$ and its other $$n-1$$ elements are permuted in the $$i$$-th way. Note that $$1 \leq i \leq (n-1)!$$. It doesn't matter how the permutations are ordered, as long as $$i \not = j \implies P_i \not = P_j$$. Now, for such a permutation $$P_i$$, create a set $$S_i$$ such that $$P_i \in S_i$$ and $$S_i$$ contains all circular shifts of $$P_i$$. For instance, if $$P_1 = (1,2,\dots ,n-1,n)$$, then

$S_1 = \{ (1,2,\dots ,n-1,n),(2,3,\dots ,n,1),\dots ,(n,1, \dots ,n-2,n-1) \}.$

Note that all pairs of elements of $$S_i$$ have the T property. If we choose $$(n-1)! + 1$$ distinct permutations of $$A$$, then by the pigeonhole principle there must be two in the same set $$S_k$$, because there are only $$(n-1)!$$ of those. Those two permutations have the T property. · 4 years ago