Order the steps involved to find whether a simple graph with 15 vertices each of

Question

Order the steps involved to find whether a simple graph with 15 vertices each of degree 5 exist. Because deg(v) is even for v e V1, the first term in the right-hand side of the last equality and the sum of the two terms on the right-hand side of the last equality are even. Hence there cannot be a graph with 15 vertices of odd degree 5. VEV ve ½ Let V1 and V2 be the set of vertices of even degree and the set of vertices of odd degree in an undirected graph with m edges Hence there can be a graph with 15 vertices of odd degree 5. Because all the terms in the sum are odd, there must be an even number of such terms and hence, there are an even number of vertices of odd degree Reset

Explanation / Answer

It starts with the "Let V1 and V2 be.." box
Then it comes to the " 2m = .." equation box
This is followed by " because degree V is even"
This is followed by " Because all the terms in the sum are odd, .."
Last step is " Hence there cannot be a graph with 15 vertices of odd degree 5"

So, the sequence of boxes is : 4 -> 3 -> 1 -> 6 -> 2

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