A permutation problem

I have got this intriguing problem from my friend.

2 distinct permutations (x1,x2,...,xn)(x_1,x_2,...,x_n) and (y1,y2,...,yn)(y_1,y_2,...,y_n) of the set A=(1,2,...,n)A=(1,2,...,n) are considered to have T property if and only if xi≠yix_i\neq y_i for all i=1,2,...,ni=1,2,...,n.

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

#Combinatorics #Permutation #MathProblem #Math

Note by Hunter Killer
7 years, 10 months ago

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4 votes

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Comments

Let PiP_i be the permutation of AA such that its first element is 11 and its other n−1n-1 elements are permuted in the ii-th way. Note that 1≤i≤(n−1)!1 \leq i \leq (n-1)!. It doesn't matter how the permutations are ordered, as long as i≠j  ⟹  Pi≠Pji \not = j \implies P_i \not = P_j. Now, for such a permutation PiP_i, create a set SiS_i such that Pi∈SiP_i \in S_i and SiS_i contains all circular shifts of PiP_i. For instance, if P1=(1,2,…,n−1,n)P_1 = (1,2,\dots ,n-1,n), then

S1={(1,2,…,n−1,n),(2,3,…,n,1),…,(n,1,…,n−2,n−1)}. 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 SiS_i have the T property. If we choose (n−1)!+1(n-1)! + 1 distinct permutations of AA, then by the pigeonhole principle there must be two in the same set SkS_k, because there are only (n−1)!(n-1)! of those. Those two permutations have the T property.

Tim Vermeulen - 7 years, 10 months ago
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