Permutations Containing and Avoiding 123 and 132 Patterns
Combinatorics
2007-05-23 v1
Abstract
We prove that the number of permutations which avoid 132-patterns and have exactly one 123-pattern equals (n-2)2^(n-3). We then give a bijection onto the set of permutations which avoid 123-patterns and have exactly one 132-pattern. Finally, we show that the number of permutations which contain exactly one 123-pattern and exactly one 132-pattern is (n-3)(n-4)2^(n-5).
Cite
@article{arxiv.math/9903169,
title = {Permutations Containing and Avoiding 123 and 132 Patterns},
author = {Aaron Robertson},
journal= {arXiv preprint arXiv:math/9903169},
year = {2007}
}
Comments
5 pages