Graph theory

Question

Let G be a graph with 10 vertices, and d(v) be the degree of a vertex v. The following conditions are holds for Graph G.

1. 3 $\leq$ d (v) $\leq$ 5 for each vertex v in G.
2. Not every vertex degree is even
3. No two odd degree vertices are of the same degree

Let X be the number of edges, Y be the vertices having even degree and Z be the vertices having odd degree in G. Find the value of X+10Y+100Z?

Answer 1

available degree of any vertex is 3,4,5 any one of them

now not all vertices are even means not all of them are 4 .. means there is either a 3 or 5  ..and there can be only one of 3 or 5 as no two odd vertices can be same here.

 noe say 9 vertices with deg 4 and 1 vertex of degree 3

then summation of degree 4*9 +3=39 but summation of degree should always even(theorem)

so only possibility is 8 vertces with 4 degree 1 with 3 and 1 with 5 .

 

and  by theorwm no of edes= (total degree/2)=(4*8+3+5=40)/2=20

edge=20 =X  vertices of even degree=8=Y  vertices of odd degree=2

X+10Y+100Z=20+80+200=300

300 should be the answer