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We prove that the number of permutations which avoid 132-patterns and have exactly one 123-pattern, equals $(n-2)2^{n-3}$, for $n \ge 3$. 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 $(n-3)(n-4)2^{n-5}$, for $n \ge 5$.

Permutations Containing and Avoiding $\textit{123}$ and $\textit{132}$ Patterns