Question about the graphical sequences

Question

A sequence $( d_1, d_2,...,d_p)$ is said to be graphic if and only if it is the degree sequence of some simple graph with p vertices. Show that the sequences $(7,5,5,5,3,2,1)$ and $(6,6,5,4,2,2,1)$ are not graphic. So for the first one 7 can't work because there are only 6 other vertices to connect to. But what if you have one degree of multiplicity 2? Then 7 works and I don't think this is what the question is looking for as an answer anyways. Using graph theory in Discrete Math, how would you solve this?

Answer 1

I think you're arguement for the first one holds. Usually in these kinds of problems graphs don't have two edges connecting the same vertices.

For the second, two $6$s means that every vertex must has at least two neighbors, since both $6$s connect to every other vertex. But then you can't have a $1$ in the sequence.

Answer 2

I think the simple graphs with no multiple edges are assumed here. So, your argument is correct for the first case, that is, the highest degree is less than the number of vertices. For the second case use the powerful Havel–Hakimi criterion: if a decreasing sequence is graphical, then the sequence obtained by deleting the first element $d$ and subtracting $1$ from the first $d$ of the rest elements must be graphical too. So, the sequence $543110$ has to be graphical but it's not the case (since a graph can not contain a vertice of the maximum degree and the isolated vertice).

Answer 3

It is impossible to draw this graph. A simple graph has no parallel edges nor any loops. There are only 7 vertices in first sequence, so each vertex can only be joined to at most six other vertices, so the maximum degree of any vertex would be 6. Hence, you can’t have a vertex of degree 7