By David Guichard

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**Extra resources for An Introduction to Combinatorics and Graph Theory**

**Sample text**

I−1, i+1, . . n} in all possible ways so that none of these n − 2 numbers is in the correct place. There are 48 Chapter 2 Inclusion-Exclusion Dn−2 ways to do this. Then, keeping 1 in position i, derange the numbers {i, 2, 3, . . , i − 1, i + 1, . . n}, with the “correct” position of i now considered to be position 1. There are Dn−1 ways to do this. Thus, Dn = (n − 1)(Dn−1 + Dn−2 ). We explore this recurrence relation a bit: Dn = nDn−1 − Dn−1 + (n − 1)Dn−2 (∗) = nDn−1 − (n − 2)(Dn−2 + Dn−3 ) + (n − 1)Dn−2 = nDn−1 − (n − 2)Dn−2 − (n − 2)Dn−3 + (n − 1)Dn−2 = nDn−1 + Dn−2 − (n − 2)Dn−3 (∗) = nDn−1 + (n − 3)(Dn−3 + Dn−4 ) − (n − 2)Dn−3 = nDn−1 + (n − 3)Dn−3 + (n − 3)Dn−4 − (n − 2)Dn−3 = nDn−1 − Dn−3 + (n − 3)Dn−4 (∗) = nDn−1 − (n − 4)(Dn−4 + Dn−5 ) + (n − 3)Dn−4 = nDn−1 − (n − 4)Dn−4 − (n − 4)Dn−5 + (n − 3)Dn−4 = nDn−1 + Dn−4 − (n − 4)Dn−5 .

There are different ways to write permutations when thought of as functions. Two typical and useful ways are as a table, and in cycle form. Consider this permutation σ: [5] → [5]: σ(1) = 3, σ(2) = 4, σ(3) = 5, σ(4) = 2, σ(5) = 1. In table form, we write this as 13 24 35 42 51 , which is somewhat more compact, as we don’t write “σ” five times. In cycle form, we write this same permutation as (1, 3, 5)(2, 4). Here (1, 3, 5) indicates that σ(1) = 3, σ(3) = 5, and σ(5) = 1, whiile (2, 4) indicates σ(2) = 4 and σ(4) = 2.

Such a permutation is called a derangement of [n]. Let S be the set of all permutations of [n] and Ai be the permutations of [n] in which n i is in the correct place. Then we want to know | i=1 Aci |. : once i is fixed in position i, the remaining n − 1 integers can be placed in any locations. What about |Ai ∩ Aj |? If both i and j are in the correct position, the remaining n − 2 integers can be placed anywhere, so |Ai ∩ Aj | = (n − 2)!. 2 Forbidden Position Permutations 47 In the same way, we see that |Ai1 ∩ Ai2 ∩ · · · ∩ Aik | = (n − k)!.