## Number of permutations to obtain identity

1. The problem statement, all variables and given/known data

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

2. Relevant equations

${n \choose 2} = \frac{n!}{k!(n-k)!}$

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 $f^{n}$ obtained by composing n successive applications of f.
Consequence: Since composition of functions is associative, we also have $f^{k} o f^{n-k}$ whenever 0 <= k <= n.

3. 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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 Recognitions: Science Advisor note that n choose 2 is (n-1)n/2. imagine that f takes 1→k. then (k-1 k)f takes 1→k-1. similarly (k-2 k-1)(k-1 k)f takes 1→k-2 after applying at most k-1 such adjacent transpostions, we arrive a product: (1 2)(2 3).....(k-1 k)f which takes 1→1. ask yourself: what the maximum value of k? now we have a permutation g that takes 1→1. suppose g takes 2→m, and repeat. won't you eventually wind up with the identity? now sum the number of adjacent transpositions used, and use a well-known formula for this sum.

 Tags combinatorics, identity, permutation, transposition