Graph $G$ has 9 vertices and 21 edges. 3 vertices have degree $x$, 3 have degree $y$, and 3 have degree $z$. Where $x,y,z$ are allowed to be the same number.
a) How many graph G could satisfy these criteria?
Total deg (G) = 3x+3y+3z = 2 * 21
3x + 3y + 3z = 42
x + y + z = 14
All the different possibles values that x, y, and z can have to satisfy x + y + z = 14 are 16 choose 14. So there are 120 G graphs that could satisfy these criteria.
b) How many graphs G could satisfy these criteria if G has no isolated vertices?
If G has no isolated vertices then,
x + y + z = 14, where x, y, and x are equal or greater than 1.
Hence, I thought that since I already know there are 120 possible G graphs (where x,y,z can be equal to 0). Then I could just count the possible ways x + y + z = 14 when x = 0 + when y = 0 + when z = 0; and then subtract that value from 120 to get the total possible ways x + y + z = 14 if x,y,z are greater than or equal to 1. My answer was 75...
Am I doing something wrong?