A restaurant offers 6 choices of appetizer, 8 choices of a main meal, and 5 choices of dessert. A customer can choose to eat just one course, or two different courses, or all three courses. Assuming all choices are available, how many different possible meals does the restaurant offer?

377 choices

Step-by-step explanation:

From the above question, we are told that

A restaurant offers 6 choices of appetizer, 8 choices of main meal and 5 choices of dessert. A customer can choose to eat just one course, or two different courses, or all three courses.

Let us represent each choice by :

A = Appetizer = 6

M = Main meal = 8

D = Dessert = 5

a) The combination of the 3 choices together

AMD=6 × 8 × 5=240

b) AM= Appetizer and Main meal

= 6 × 8 = 48

c) AD= Appetizer and Dessert

= 6 × 5 = 30

d) MD = Main meal × Dessert

= 8 × 5 = 40

e) A,M,D (each alone)=

Appetizer + Main meal + Dessert

= 6 + 8 + 5

= 19

Assuming all choices are available, how many different possible meals does the restaurant offer?

This is calculated as:

AMD + AM + AD + MD + A,M,D

240 + 48 + 30 + 40 + 19

= 377 choices

well there are two answers for that.

1: you need 120 ones then that will equal up to 120.

2: you could also do one 100 and 2 tens.

those are the answers that i have thought of.