Genius! The $$n=2$$ orbit can have $$8$$ electrons. So, we need to finish the $$n=2$$ orbit before we go to the $$n=3$$ orbit. Therefore, the configuration: $$2$$ in $$n=1$$, $$6$$ in $$n=2$$ and $$4$$ in $$n=3$$ is invalid.
There is another rule, Rule 2, to how electrons get distributed. It says that the outermost orbit cannot have more than $$8$$ electrons. For example, consider Potassium. It has 19 electrons.
In $$n=1$$ and $$n=2$$, it can have $$2$$ and $$8$$ electrons. So, we are left with $$9$$ electrons.
The $$n=3$$ orbit can have $$18$$ electrons. But, because the outermost orbit cannot have more than $$8$$ electrons, the $$n=3$$ orbit has $$8$$ electrons and the $$n=4$$ orbit has $$1$$ electron.