How to prove that total number of non-isomorphic labelled trees of order $n$ is $n^{n-2}$?

Question

I predicted the formula by finding total number of non-isomorphic labelled trees of order 1 is 1 , order 2 is 1,order 3 is 3,order 4 is 16,order 5 is 125.But how do i prove it ? I am beginner in graph theory so i will be very thankful if i get some simple approach to solve the problem(Not by using some difficult theorems).

To get the right superscript to show up, use accolades, like so: $n^{n-2}$ will typeset $n^{n-2}$. — Mees de Vries, Feb 26 '19 at 12:58
There's a proof of this in a book by Arthur Engel, "problem solving strategies". I have the copy at home, and I'm not at home now. — Sarvesh Ravichandran Iyer, Feb 26 '19 at 12:59

score 2 · Accepted Answer · answered Feb 26 '19 at 13:04

2

One approach is to use Prüfer sequences; there is a bijection between labelled trees and length-$n-2$ sequences of numbers in $[1,n]$ (the Wikipedia article gives a good description of how to go from sequence to tree and back). Since there are $n^{n-2}$ such sequences, there are also this many labelled trees.

answered Feb 26 '19 at 13:04

Parcly Taxel

105,904

Wikipedia shows how, given a labelled tree $T$, to construct a Prüfer sequence for $T$. We may define Prüfer sequences so. So unequal Prüfer sequences can't have isomorphic labelled trees. Moreover, non-isomorphic labelled trees can't have equal Prüfer sequences. Suppose not; let $T_1, T_2$ be an example minimising $n$. There are labels $v$ and $w$ such that, in each case, the algorithm removes edge $vw$. By minimality of $n$, the respective results of removing $vw$ from each tree are isomorphic to some labelled tree $T_3$. So $T_1$ and $T_2$ are both isomorphic to $T_3+uv$. – Rosie F Feb 27 '19 at 11:10

How to prove that total number of non-isomorphic labelled trees of order $n$ is $n^{n-2}$?

1 Answers1

Linked

Related