I have $5$ veggie items and $4$ meat items. How many different plates can I make? 
Janet must select three different items for each dinner she will serve. The items are to be chosen from among $5$ different vegetarian and $4$ different meat selections. If at least one of the selections must be vegetarian, 
  how many different dinners could Jane create?

The answer is $80$, I came up with a solution but I don't think it's the right way to think about the problem, is there a correct way to essentially do it?
One of the plates has to be a veggie product, so suppose to we have only have two plated with $4$ veggie options and $4$ meat options, then we have a total of $16$ possible combinations.
Thus, we introduce the $3^{rd}$ plate, and because this plate has to be veggie, and out of $5$ possible veggie options, we have then $16\times 5 = 80$ different plates. Any other ways to think about it?
 A: It is simple. We can choose any $3$ item out of all the $5+4=9$ dishes in $\color{red}{{9\choose3}}=\color{blue}{84}$ ways.
Then, consider the number dishes, that there are no vegetarian dishes, i.e. only non-veg dishes. That can be done in $\color{red}{{4\choose 3}}=\color{blue}{4}$ ways.
Then, the number of combinations of dishes that contains at least one veg dishes is $$\color{red}{{9\choose3}-{4\choose 3}}=\color{blue}{84-4}=\color{navy}{\fbox{80}}$$

I am editing this for @ Jhon's satisfaction.
If, from first I had counted the veg dishes, then the approach is the following.
First, I consider that there were $1$ veg and $2$ non-veg dishes. It can happen in $$\color{red}{{5\choose1}{4\choose2}}=\color{green}{5\times6}=\color{blue}{30}\tag 1$$ ways.  
If, there were $2$ veg and $1$ non-veg dishes, then this can happen in $$\color{red}{{5\choose2}{4\choose1}}=\color{green}{10\times4}=\color{blue}{40}\tag2$$ ways.
If, there were all $3$ veg dishes, then, it could happen in $$\color{red}{{5\choose3}{4\choose0}}=\color{green}{10\times 1}=\color{blue}{10}\tag3$$ ways.
As, you see, there are no other cases. Hence our desired result is $$\color{blue}{30+40+10}=\color{navy}{\fbox{80}}$$ 
