What is wrong with my method of computing combinations for this GRE question?

Question

Question: Jane must select 3 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 can Jane create?

Correct answer: 80

I calculated this question incorrectly using the following logic: Jane must always have at least one vegetable dish, so let's select that first, 5C1, now she can select any combination of 2 items from the remaining 8 items, 8C2, therefore the answer should be (5C1)*(8C2) = 140.

I see that I am counting some combinations twice because the two sets intersect in the veggies. However I am looking for a general unified formula to compute problems of this nature rather than the suggested solution of computing "how many all veggie dishes, 2 veggie, 1 veggie dishes" and summing. That method isn't scalable as the number of ingredients goes up.

Answer 1

You've correctly identified the flaw in your reasoning, which is that you are possibly double counting.

Here's an alternative approach: Count the total number of dinners with no restriction, which will be $9 \choose 3$, and subtract off the number of disallowed dinners, which will be those that are all meat, namely $4\choose 3$. The answer is then $84 - 4=80$.

Answer 2

We are going to pick 3 items among the 9 items (5 veg and 4 meat).

Number of selection which exactly contains one vegetarian = $5C1×4C2=30$

Number of selection which exactly contains two vegetarians = $5C2×4C1=40$

Number of selection which exactly contains three vegetarians = $5C3×4C0=10$

So Totally $30+40+10=80$