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n ΣI(n,k)x² = 1(1+x)(1+x+x²)...(1 + x + +xn-1) k=0 ... We often denote [m]x = 1+x+...+xm-1, and denote the value on the right hand side [n]!, the x-analogue of n! (this is because setting x = 1 recovers n!).

n ΣI(n,k)x² = 1(1+x)(1+x+x²)...(1 + x + +xn-1) k=0 ... We often denote [m]x = 1+x+...+xm-1, and denote the value on the right hand side [n]!, the x-analogue of n! (this is because setting x = 1 recovers n!).

Computer Networking: A Top-Down Approach (7th Edition)
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Discrete Mathematics Proving identity Problem

A pair of i < j such that  σi > σj  is an inversion of a permutation.

Let the number of permutations in Sn with k inversions be denoted by I(n, k),

The task is to Prove the identity below 

n ΣI(n, k)æk = 1(1+x)(1+x+x²)...(1 + x + ... + x¹−¹) k=0 We often denote [m]x 1 = 1+x+ +xm-¹, and denote the value on the right hand side [n]!, the x-analogue of n! (this is because setting x = 1 recovers n!).
Transcribed Image Text:n ΣI(n, k)æk = 1(1+x)(1+x+x²)...(1 + x + ... + x¹−¹) k=0 We often denote [m]x 1 = 1+x+ +xm-¹, and denote the value on the right hand side [n]!, the x-analogue of n! (this is because setting x = 1 recovers n!).
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