question about simple graph. I know that simple graph has no parellel edges and loops. My question is that I have to draw the graph on six vertices with degree sequence $(3,3,5,5,5,5)$. I draw the the graph with the given degree sequence and everytime I got the graph which is not simple. Does there exist a simple graph with these degree sequence? If no why?
 A: By the Havel-Hakimi Theorem, a descending degree sequence $\{a_1, \cdots ,\ a_k\}$ is graphical if and only if $a_1 \le k-1$ and $\{a_2 - 1, \cdots, a_{a_1+1}-1, a_{a_1 +2}, \cdots, a_k\}$ is graphical.
Applying the theorem to your case, we have
$$\{5,5,5,5,3,3\} \iff \{4,4,4,2,2\}\iff\{3,3,1,1\}\iff\{2,0,0\}$$
The latter graph is evidently impossible.
A: No fancy theoretical machinery is required.
Call the vertices $v_1,v_2,v_3,v_4,v_5$, and $v_6$. Let $v_1,v_2,v_3$, and $v_4$ be the vertices of degree $5$. There are only six vertices, so each of these vertices must be connected by an edge to every other vertex in the graph. For instance, $v_2$ must be connected to $v_1,v_3,v_4,v_5$, and $v_6$. This means that $v_5$ and $v_6$ are both connected to each of the vertices $v_1,v_2,v_3$, and $v_4$ and must therefore have degree at least $4$: the degree sequence $(3,3,5,5,5,5)$ is impossible for a simple graph.
If you’d like to explore further, this kind of reasoning leads to the easy half of the Erdős-Gallai theorem, which gives a simple computational criterion for whether a sequence is the degree sequence of some simple graph. This kind of reasoning shows that a sequence that fails the criterion cannot possibly be a degree sequence; proving that every sequence that satisfies it really is the degree sequence of some simple graph is harder. See also the comments here.
