Answer:
Therefore, Jane can create 80 different dinners.
Step-by-step explanation:
We know that Jane must select three different items for each dinner she will serve. If at least one of the selections must be vegetarian.
The items are to be chosen from among five different vegetarian and four different meat selections.
First we count the number of combinations for one vegetarian dinner and 2 meat dinners.
[tex]C_1^5\cdot C_2^4=5\cdot \frac{4!}{2!(4-2)!}=5\cdot 6=30[/tex]
Now we count the number of combinations for 2 vegetarian dinner and 1 meat dinners.
[tex]C_2^5\cdot C_1^4=10\cdot 4=40\\[/tex]
Now we count the number of combinations for 3 vegetarian dinner.
[tex]C_3^5=\frac{5!}{3!(5-3)!}=10\\[/tex]
We get 30+40+10=80.
Therefore, Jane can create 80 different dinners.
