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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 ≥ 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 ≥ 5.fatcat:xjyd6nttb5fkteeqouzpcfwic4