8 votes 8 votes A simple graph ( a graph without parallel edge or loops) with $n$ vertices and $k$ components can have at most $n$ edges $n-k$ edges $(n-k) (n-k+1)$ edges $(n-k) (n-k+1)/2$ edges Graph Theory isro2009 graph-theory graph-connectivity + – go_editor asked Jun 15, 2016 edited Dec 8, 2022 by Lakshman Bhaiya go_editor 3.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
8 votes 8 votes Ans D) graph with n vertices and k component has (n-k) ≤ e ≤(n-k)(n-k+1)/2 edges srestha answered Jun 15, 2016 srestha comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Answer D) https://gateoverflow.in/510/gate1991_01-xv Kapil answered Jun 15, 2016 Kapil comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Take an example of a complete graph on 4 vertices, no. of components =1 and no. of edges =6 which is satisfied by option D. Jhaiyam answered Aug 26, 2020 Jhaiyam comment Share Follow See all 0 reply Please log in or register to add a comment.