A restaurant offers a​ $12 dinner special that has 6 choices for an​ appetizer, 12 choices for an​ entrée, and 3 choices for a dessert. How many different meals are available when you select an​ appetizer, an​ entrée, and a​ dessert?

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.