Respuesta :
Answer: The required number of different meals is 216.
Step-by-step explanation: Given that a restaurant offers a $12 dinner special that has 6 choices for an appetizer, 12 choices for an entree and 3 choices for a dessert.
We are to find the number of different meals that are available when we select an appetizer, an entree and a dessert.
The number of ways in which one appetizer can be selected from 6 choices of appetizers is
[tex]n_1=^6C_1=\dfrac{6!}{1!(6-1)!}=\dfrac{6\times5!}{1\times5!}=6,[/tex]
the number of ways in which one entree can be selected from 12 choices of entrees is
[tex]n_2=^{12}C_1=\dfrac{12!}{1!(12-1)!}=\dfrac{12\times11!}{1\times11!}=12[/tex]
and the number of ways in which one dessert can be selected from 3 choices of desserts is
[tex]n_3=^3C_1=\dfrac{3!}{1!(3-1)!}=\dfrac{3\times2!}{1\times2!}=3.[/tex]
Therefore, the number of different meals that can be available is given by
[tex]n=n_1\times n_2\times n_3=6\times12\times3=216.[/tex]
Thus, the required number of different meals is 216.
