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Defining an element σ ∈ Sn is equivalent to specifying the list σ(1), σ(2), . . , σ(n) consisting of the n numbers 1, 2, . . , n taken in some order with no repetitions. To do this we have • n choices for σ(1), • n − 1 choices for σ(2) (taken from the remaining n − 1 elements), • and so on. In all, this gives n × (n − 1) × · · · × 2 × 1 = n! choices for σ, so |Sn | = n! as claimed. We will often describe σ using the notation σ= 1 2 ... n . σ(1) σ(2) . . 6. The elements of S3 are the following, ι= 1 2 3 , 1 2 3 1 2 3 , 2 3 1 1 2 3 , 3 1 2 1 2 3 , 1 3 2 1 2 3 3 2 1 1 2 3 .

2 3 1 Often it is useful to display the effect of a permutation σ : X −→ X by indicating where each element is sent by σ with the aid of arrows. To do this we display the elements of X in two similar rows with an arrow joining xi in the first row to σ(xi ) in the second. For example, A B C the permutation σ = acting on X = {A, B, C} can be displayed as B C A A@ @ A B @ oo C @@ o@@oo @@ooooo @@@ @ oo@ wooo B C 3. THE SIGN OF A PERMUTATION 31 We can compose permutations by composing the arrows.

The next result is a formal version of what is usually called the Pigeonhole Principle. 3 (Pigeonhole Principle: first version). a) If there is an injection m −→ n then m n. b) If there is a surjection m −→ n then m n. c) If there is a bijection m −→ n then m = n. Proof. a) We will prove this by Induction on n. Consider the statement P (n) : For m ∈ N0 , if there is a injection m −→ n then m n. When n = 0, there is exactly one function ∅ −→ ∅ (the identity function) and this is a bijection; if m > 0 then there are no functions m −→ ∅.