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0 $begingroup$ This question already has an answer here: How many permutations of ${1, ldots, n}$ exist such that none of them contain $(i, i+1)$ (as a sequence) for $i in {1,…,(n-1)}$? 3 answers If $6$ people are standing up in queue for a picture, then in how many ways can they be re queued for the picture if no person has same right hand neighbor? Why is $6!/2$ wrong? (The answer is $309$ .) Edit : Based on @Phicar's suggestion in the comments, an attempt using the Inclusion-Exclusion Principle: $$6!- 5.5! +4.4! -3 .3! + 2.2! - 1$$ combinatorics permutations inclusion-exclusion share | cit