Apr 17, 2016 · Remember one thing , AND means multiply and OR means add (For eg. if you had said to choose either one main course OR one dessert, then we would have added the selections) For 1. , selections = 2C1 = 2. For 2. , selections = 4C1 = 4. For 3. , selections = 3C1 = 3. Total possibilites = 2C1 × 4C1 × 3C1 = 2 × 4 ×3 = 24.
There are 3 appetizers, 8 main dishes, and 5 desserts on a restaurant menu. (a) In how many different ways one can choose a full dinner course in this restaurant? (b) On Friday night the appetizer and main dish menus are mixed and has a total of 11 choices. One can pick one for the appetizer and a different one for the main dish. How many different
Question 244464: if a menu has a choice of 4 appetizers, 4 main courses, and 3 desserts, then the sample space for all possible dinners has how many outcomes? A) 48 B) 16 C) 28 D) 3 Answer by checkley77(12844) (Show Source):
A. 32 B. 19 C. 60 D. 3; Question: If a menu has a choice of 4 appetizers, 5 main courses, and 3 desserts, how many dinners are possible if each includes one appetizer, one main course, and one dessert. A. 32 B. 19 C. 60 D. 3
Expert Answer Thus, 504 different means are available to select an appetizer, an entrée, and a dessert.
You can pick 2 appetizers out of 5 in 10 ways. You can pick 4 main dishes out of 11 in 330 ways. You can pick 2 desserts out of 6 in 15 ways. 10 x 330 x 15 = 49,500 different combinations.
2· 6· 4· 3 = 144 possible meals. Example: David has 6 books that he wishes to arrange on his desk. How many different arrangements are possible?
5*3*4 = 60, so 60 different main meals could be ordered.
4 course meal: A 4 course dinner menu includes an hors d'oeuvre, appetizer, main course, and dessert.
720Permutation: Listing your 3 favorite desserts, in order, from a menu of 10. P(10,3) = 720.
We see that there are 18 different three-course meals. However, there is an easier way to get at this answer. Notice that in order to choose a three-course meal, we need to make three decisions: 1.
Since you make independent choices, simply multiply the number of options for each course together, which gives 4×8×3=96. As JWeissman states, the total number of three course meals, assuming each includes one starter, one main, and one dessert, would be 4⋅8⋅3=96. This uses the fundamental principle of counting.Sep 26, 2015
Sometimes we want to know how many different combinations can be made of a variety of items. The fundamental counting principle states that the number of ways in which multiple events can occur can be determined by multiplying the number of possible outcomes for each event together.Apr 25, 2013
If a diner can select from 4 types of soup. 3 kinds of sandwiches, 5 desserts and 4 drinks, they can have 240 possible different combinations!