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lizarton

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## Homework Statement

Let [itex]s*(f)[/itex] be the minimum number of transpositions of adjacent elements needed to transform the permutation [itex]f[/itex] to the identity permutation. Prove that the maximum value of [itex]s*(f)[/itex] over permutations of [itex][n][/itex] is [itex]{n \choose 2}[/itex]. Explain how to determine [itex]s*(f)[/itex] by examining [itex]f[/itex].

## Homework Equations

[itex]{n \choose 2} = \frac{n!}{k!(n-k)!}[/itex]

Perhaps...

Definition: The identity permutation of [n] is the identity function from [n] to [n]; its word form is 1 2 ... n. A transposition of two elements in a permutation switches their positions in the word form.

A permutation f of [n] is even when P(f) is positive, and it is odd when P(f) is negative. When n = 1, there is one even permutation of [n] and no odd permutation. For n >= 2, there are n!/2 even permuatations and n!/2 odd permutations.

My book solves a problem that counts the number of exchanges of entries needed to sort the numbers into the order 1,2,...,n (given a list of n numbers 1 through n). The solution defines the nth iterate of f: A --> A.

Definition: The nth iterate of f: A --> A is the function [itex]f^{n}[/itex] obtained by composing n successive applications of f.

Consequence: Since composition of functions is associative, we also have [itex]f^{k} o f^{n-k}[/itex] whenever 0 <= k <= n.

## The Attempt at a Solution

The number of subsets of length 2 of a permutation of length n is n choose 2. I can see that from the definition, but how does one formulate a formal proof from just the definition?

I can guess that the scenario requiring the most trials is when f is the permutation n...2 1.

Any help would be greatly appreciated!

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