5. Let G be the graph on n vertices obtained from Kn by removing a single edge. Prove that the c

First, we know that the chromatic polynomial of a complete graph $K_n$ is given by $P(K_n, k) =

5. Let G be the graph on n vertices obtained from Kn by...

Transcript

00:01 In this question we have been given that let g be the graph on n vertices the number of vertices in g it is given to be n it is obtained from kn that is complete graph by removing a single edge okay so now the number of edge in g will be what complete graph has number of edges equal to this minus one is a number of edge in g we have to prove that the chromatic polynomial for this graph g it is going to look like k times k minus one so on up till k minus n plus three and k minus n plus two whole square so let us see how we are going to prove this so first of all let's see what is a chromatic polynomial it is a special function which describes the number of ways we find a proper coloring okay so i will write down so it gives the number of ways we can find a proper coloring of a graph we can find a proper coloring of the graph with given let's say we have given some m colors so this is how we define the chromatic polynomial so now let us see how we are going to prove this so for that we are going to consider the complete graph so consider a complete graph kn having n vertices just take a random vertex and then it can be colored by k ways okay so for any random vertex it can be colored in k ways it can be colored in a number of color let's say it is k okay so it will be colored in k ways if we take another vertex then it can be colored by only k minus one ways okay because we are doing proper coloring so if we take another vertex it can be colored in k minus one ways as is it adjacent to other n minus one vertices so it can't have the same color so we are getting k minus one choices okay hence we can say that we can say that the complete graph kn can be colored in can be colored in how many ways so for the first vertex we have k choices next one k minus one then again next one it is k minus two and the nth vertex will be having k minus n let this is for the nth vertex number of ways will be k minus n plus one so this is the total number of ways to color a complete graph now what i will what i am going to consider g which is formed by removing one edge from kn correct we will let e equals to ub now we have to find out this is an edge in g now what we are going to do the number of k colorings of g k colorings of g in which u and v have different color is the same with or without h e with or without h e and therefore it is equals to and therefore we say that total number of k coloring will be it is equal to p kn of k okay so that is a chromatic polynomial basically similarly the number of k colorings in g k coloring in g in which u and v have the same color u and v have the same color does not change…

Question

5. Let G be the graph on n vertices obtained from Kn by removing a single edge. Prove that the chromatic polynomial of G is k(k - 1) ... (k - n + 3)(k - n + 2)^2.

Question

5. Let G be the graph on n vertices obtained from Kn by removing a single edge. Prove that the chromatic polynomial of G is k(k - 1) ... (k - n + 3)(k - n + 2)^2.

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