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Lecture 20
Andrei Antonenko
March 26, 2003
1 Permutations
Let we have n numbers from 1 to n: 1, 2 ,... , n. If we change their order we get their per- mutation. We will write these numbers in brackets, for example, (34152) is a permutation of numbers from 1 to 5.
Example 1.1. There are 6 different permutations of numbers 1 , 2 and 3 :
(123) (132) (213) (231) (312) (321) Let’s compute the total number of permutations of numbers from 1 to n. On the first place in the permutation we can put any of n elements. On the second place we can put any element which is not equal to the element on the first place — so we have (n − 1) possibilities; than to the third place we can put any element which is not on the first and the second place, so totally we have (n − 2) possibilities, etc. Finally, we will have only one possibility to put an element on the last n-th place. So, total number of permutations of n elements is n(n−1)(n−2) · · · 2 ·1 = n! (By definition, n! is a product of all numbers from 1 to n. It is called n-factorial). So, we see that the number of permutations increases very fast. We have just 3! = 6 different permutations of numbers from 1 to 3, 4! = 24 different permutations of numbers from 1 to 4, 5! = 120 different permutations of numbers from 1 to 5, and the number of permutations of the numbers from 1 to 10 is already equal to 10! = 3628800. We will use the following notation. The permutations will be denoted by Greek letters σ and τ — “sigma” and “tau” respectively. For example, we will write σ = (3241) and τ = (231). The set of all permutations from 1 to n will be denoted by Sn. For example, as we saw already, S 3 = {(123), (132), (213), (231), (312), (321)}. By σ(i) we will denote the i-th element of the permutation σ. For example, if σ = (3241), then σ(1) = 3, σ(2) = 2, σ(3) = 4, and σ(4) = 1.
Definition 1.2. Two elements of the permutation form an inversion if the largest stands to the left of the smallest.
Definition 1.3. The permutation is called even if the total number of inversions is even, and odd if the total number of inversions is odd. The number (−1)# of inversions^ is called the sign of the permutation, and is equal to 1 for even permutations and -1 for odd. It is denoted by sgn(σ).
Example 1.4. • σ = (123). There are no inversions at all. So, it is even, and sgn(σ) = 1.
- σ = (132). Numbers 3 and 2 form an inversion. So, there is only 1 inversion. So, it is add, and sgn(σ) = − 1.
- σ = (213). Numbers 2 and 1 form an inversion. So, there is only one inversion. So, it is odd, and sgn(σ) = − 1.
- σ = (231). Numbers 2 and 1 form an inversion. Numbers 3 and 1 form an inversion. So, totally there are 2 inversions. So, it is even, and sgn(σ) = 1.
- σ = (312). Numbers 3 and 1 form an inversion. Numbers 3 and 2 form an inversion. So, totally there are 2 inversions. So, it is even, and sgn(σ) = 1.
- σ = (321). Numbers 3 and 2 form an inversion. Numbers 3 and 1 form an inversion. Numbers 2 and 1 form an inversion. So, there are 3 inversions. So, it is odd, and sgn(σ) = − 1.
The main fact about permutations is the following.
Lemma 1.5. If we interchange any 2 elements in the permutation, its sign changes, i.e. if it was even, it becomes odd, and other way round.
Example 1.6. Let σ = (25431). The following pairs of numbers form an inversion: 21, 54, 53, 51, 43, 41, 31. So, we have 7 inversions, and this permutation is odd. Let’s interchange 5 and 1. We’ll get τ = (21435). The following pairs of numbers form an inversion: 21, 43. So, we have just 2 inversions, so this permutation is even.
Proof of the lemma. First let’s note, that if we transpose 2 consecutive elements, the number of inversions changes exactly by 1. So, the parity of the permutation changes. Transposition of the elements i and j can be done by 2s + 1 consecutive transpositions of the adjacent elements: first interchange i with all elements on its way to j, and than move j to the place where we had i. So, the sign will change odd number of times, and the parity of the permutation will change.
