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Electronic Notes in Discrete Mathematics 38 (2011) 147–148 www.elsevier.com/locate/endm
Complete Intersection Problems for Finite Permutations Vladimir Blinovsky Dobrushin Math. Lab. Institute of Information Transmission Problems, Moscow Russia
[email protected]
Abstract We find the maximal number of permutations on a set of n elements such that any pair of permutations has at least t common cycles. Keywords: Complete intersection theorem, permutations
1
Main result
Recall notation: [n] = {1, 2, . . . , n}, 2[n] = {A ⊂ [n]}, and [n] the standard = {A ∈ 2[n] : |A| = i}. Denote by Σ(n) the set of all permutations on the i n (−1)i be the number of permutations set [n], |Σ(n)| = n!. Let also f (n) = n! i! i=0
on the set [n] that do not have cycles of length 1. One can easily show that (1)
n!/e − 1 < f (n) < n!/e + 1.
Each permutation σ(n) ∈ Σ(n) is determined by its cycles σ(n) = {s1 , . . . , sp }. We say that a pair of permutations σ1 (n), σ2 (n) has intersection t if this pair has at least t common cycles. The problem that we solve in the present paper 1571-0653/$ – see front matter © 2011 Elsevier B.V. All rights reserved. doi:10.1016/j.endm.2011.09.025
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V. Blinovsky / Electronic Notes in Discrete Mathematics 38 (2011) 147–148
is finding the maximal number N(n, t) of permutations on the set [n] with a given pairwise intersection t for any 1 ≤ t ≤ n. The main result of this work is proving the following theorem. Theorem 1.1 Let t ≥ 2, and let = t + 2r be the largest number not greater than n satisfying the relation n− +t f n − 2 − i n− i −t ≤ n−+1i=0 (2) . n−+1 2( − 1) +t f n− 2 +1−i i i=0
Then (3)
N(n, t) =
t+2r i=t+r
t + 2r i
n−t−2r j=0
n − t − 2r f (n − i − j). j
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