The number of derangements ( !n ) of ( n ) objects is [ !n = n!\sum_{k=0}^{n}\dfrac{(-1)^k}{k!}. ] For ( n=4 ), ( !4 = 9 ).
Accedmychevron_right11thchevron_rightmathchevron_rightPermutations And Combinationschevron_rightExercise 7.1
The number of derangements ( !n ) of ( n ) objects is [ !n = n!\sum_{k=0}^{n}\dfrac{(-1)^k}{k!}. ] For ( n=4 ), ( !4 = 9 ).