I have constructed two polyhedrons as follows: There are $\binom{6}{3}$ triangles and $\binom{6}{2}$ squares. Every triangle is connected via an edge with $3$ distinct squares (one for each vertex of the triangle), every square is connected in the same way with $4$ distinct triangles. This construction gives a non planar graph. The result looks as follows:

I calculated its Euler characteristic by using that there are aside from the triangles and squares exactly $\binom{6}{3}$ dodecagons as faces, because one edge each from 3 distinct triangles, one edge each from 3 distinct squares and 6 edges between triangles and squares gives a face, and, then, using $v=3\binom{6}{3}+4\binom{6}{2}$, $e=6\binom{6}{3}+4\binom{6}{2}$ and $f=2\binom{6}{3}+\binom{6}{2}+1$ for the genus $g$: $$2-2g=\chi = v-e+f=\binom{6}{2}-\binom{6}{3}+1=-4$$ and, consequently, the genus as $3$.

But if I now consider the same construction but now with $\binom{7}{3}$ triangles and $\binom{7}{2}$ pentagons obtaining now $\binom{7}{3}$ dodecagons, I can still build the graph and build it comutationally.

But I get, then, by $v=3\binom{7}{3}+5\binom{7}{2}$, $e=6\binom{7}{3}+5\binom{7}{2}$ and $f=2\binom{7}{3}+\binom{7}{2}+1$ the Euler characteristic $\chi = v-e+f=-13$. This gives me a genus which is a fraction. Where am I going wrong? Does the formula $\chi=2-2g$ not work without restrictions? Or is my counting off? The $+1$ in my calculation for the faces comes from the "outside faces"

EDIT: Using labels in the first case, I find the following net, which gives the graph by identification of identical squares/triangles. Note that the edges between squares and triangles are contracted. The dodecagons become, therefore, hexagons. The labels are chosen according to the construction described hereinabove, representing in particular the contracted edges.

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