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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
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.


Source: Aguiar, Marcelo - Department of Mathematics, Texas A&M University


Collections: Mathematics