Math 431, Section 500 Solutions to Exam II Summary: Math 431, Section 500 Solutions to Exam II 1. Find a formula for the number of permutations of [n] in which exactly k integers are in their natural positions. Solution. To give such a permutation we may first choose k integers to be left in their natural positions (this can be done in n k ways) and then permute the remaining n-k integers in such a way that no element is left in its natural position (this is the same as choosing a derangement of n - k elements, which can be done in Dn-k ways). By the multiplication principle, the answer is n k · Dn-k . 2. Find the number of permutations a1a2a3a4 of [4] such that ai = i, i + 1 i = 1, 2, 3, 4 . Solution. Let be the set of all permutations of [4] and for each i = 1, 2, 3, 4 let Ai denote the set of permutations a1a2a3a4 such that ai = i or ai = i + 1. We want #\(A1 A2 A3 A4). We compute this cardinality with the Principle of Inclusion-Exclusion. Note that A4 is special, as the only relevant restriction in this case is a4 = 4. Collections: Mathematics