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.

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## Comments

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TopNewestLet \(P_i\) 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.

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